Answer:
The low tide, when it is 10 feet below the high tide would be at 7am and 7pm
Step-by-step explanation:
There are 2 new seats in each row in a school auditorium. There are 20 rows in the auditorium. Each new seat costs $84 what is the cost for the new seats
A miner is trapped in a mine containing 3 doors:1. The first door leads to a tunnel that will take him to safety after 3 hours of travel.2. The second door leads to a tunnel that will return him to the mine after 5 hours of travel.3. The third door leads to a tunnel that will return him to the mine after 7 hours.If we assume that the miner is at all times equally likely to choose any one of the doors (supposing themine shaft is so disorienting that he cannot tell which door he chose before), LetXdenote the lengthof time until the miner reaches safety.Compute the variance, Var(X).
Answer:
The miner should take door number 1
The number of people attending graduate school at a university may be
modeled by the quadratic regression equation y = 10x2 - 40x+8, where x
represents the year. Based on the regression equation, which year is the best
prediction for when 2528 people will attend graduate school?
O A. Year 15
B. Year 23
O C. Year 18
D. Year 20
Answer:
It’s C
Step-by-step explanation:
But thanks for helping me get it wrong toxo360❤️
In the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.
What is the quadratic equation?The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of: ax² + bx + c = 0.
The number of people attending graduate school at a university may be modeled by the quadratic regression equation y = 10x2 - 40x+8, where x represents the year.
The regression equation, which year is the best prediction for when 2528 people will attend graduate school is;
[tex]\rm y = 10x^2 - 40x+8\\\\2528 = 10x^2 - 40x+8\\\\ 10x^2 - 40x+8\\\\ 10x^2 - 40x+8-2528=0 \\\\ 10x^2 - 40x-2520=0\\\\\text{Divide by 10 both sides}\\\\x^2-4x-252=0\\\\x^2-18x+14x-252=0\\\\x(x-18)+14(x-18)=0\\\\(x-18)(x+14)=0\\\\x-18=0, \ x=18\\\\x+14=0, \ x=-14[/tex]
Hence, the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.
Learn more about quadratic equations here;
https://brainly.com/question/24298861
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5. A meteorologist measured the average rainfall received in cities A and
B. Both cities received 11 inches of rainfall in total. While City A received x
inches of rain, City B experienced three times the amount of rainfall than
City A. Find the number of inches of rain City A received.
Answer:
2.75 inches
Step-by-step explanation:
City A received x inches of rain.City B experienced three times the amount of rainfall than City A, therefore :
City B received 3x inches of rain.Since both cities received 11 inches of rainfall in total.
We have that:
x+3x=11
4x=11
Divide both sides by 4
x=2.75 Inches
Therefore, we City A received 2.75 inches of rain.
Eric’s average income for the first 4 months of the year is $1,450.25, what must be his average income for the remaining 8 months so that his average income for the year is $1,780.75?
Answer: $
Answer:
$1946
Step-by-step explanation:
His average for the first four months
= $1,450.25
So he earned 4*$1,450.25
= $5801 for the first four months.
Then for the year his average is $1,780.75.
So he earned $1,780.75*12 for the year.
=$ 21369
So the amount he earned for the remaining 8 months was 21369-5801
= $15568
The average for the 8 Months= 15568/8
= $1946
Solve this inequality.
-35 – 2 >7
A.
B.
C.
D. > -3
Answer:
-2> 7+35
-2>42
0<21
the answer is 0 is less than 21
Lockheed Martin, the defense contractor designs and build communication satellite systems to be used by the U.S. military. Because of the very high cost the company performs numerous test on every component. These test tend to extend the component assembly time. Suppose the time required to construct and test (called build time) a particular component is thought to be normally distributed, with a mean equal to 60 hours and a standard deviation equal to 9.4 hours. To keep the assembly flow moving on schedule, this component needs to have a build time between 52 and 70 hours. Find the probability that the build time will be such that assembly will stay on schedule.
Answer:
[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]
And we can find this probability with this difference:
[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]
And if we use the normal standard distribution or excel we got:
[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]
Step-by-step explanation:
Let X the random variable that represent the time required to construct and test a particular component of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(60,9.4)[/tex]
Where [tex]\mu=60[/tex] and [tex]\sigma=9.4[/tex]
We want to find this probability:
[tex]P(52<X<70)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(52<X<70)=P(\frac{52-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{70-\mu}{\sigma})=P(\frac{52-60}{9.4}<Z<\frac{70-60}{9.4})=P(-0.851<z<1.064)[/tex]
And we can find this probability with this difference:
[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)[/tex]
And if we use the normal standard distribution or excel we got:
[tex]P(-0.851<z<1.064)=P(z<1.064)-P(z<-0.851)=0.856-0.197=0.659[/tex]
This table gives a few (x,y) pairs of a line in the coordinate plane.
Answer:
-26
Step-by-step explanation:
y=mx+c
m= (y2 - y1 ) / (x2-x1)
m= (10-1) / (48 -36)
m = 0.75
10 = 0.75(48) + c
c = 10 - 36
c = -26
An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch. What is the probability that a ball bearing is:___________.
a. between the target and the actual mean?
b. between the lower specification limit and the target?
c. above the upper specification limit?d. below the lower specification limit?
Answer:
(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.
(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.
(c) Probability that a ball bearing is above the upper specification limit is 0.0401.
(d) Probability that a ball bearing is below the lower specification limit is 0.0006.
Step-by-step explanation:
We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.
Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.
Let X = diameter of the ball bearings
SO, X ~ Normal([tex]\mu=0.753,\sigma^{2} =0.004^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 0.753 inch
[tex]\sigma[/tex] = standard deviation = 0.004 inch
(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X [tex]\leq[/tex] 0.75 inch)
P(X < 0.753) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.753-0.753}{0.004} } }[/tex] ) = P(Z < 0) = 0.50
P(X [tex]\leq[/tex] 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.75) = 1 - P(Z < 0.75)
= 1 - 0.7734 = 0.2266
The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.
Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.
(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X [tex]\leq[/tex] 0.74 inch)
P(X < 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z < -0.75) = 1 - P(Z [tex]\leq[/tex] 0.75)
= 1 - 0.7734 = 0.2266
P(X [tex]\leq[/tex] 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -3.25) = 1 - P(Z < 3.25)
= 1 - 0.9994 = 0.0006
The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.
Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.
(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)
P(X > 0.76) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] > [tex]\frac{0.76-0.753}{0.004} } }[/tex] ) = P(Z > -1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)
= 1 - 0.95994 = 0.0401
The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.
(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)
P(X < 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z < -3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)
= 1 - 0.9994 = 0.0006
The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.
Using the normal distribution, it is found that there is a
a) 0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.
b) 0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.
c) 0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.
d) 0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.In this problem:
Mean of 0.753 inch, hence [tex]\mu = 0.753[/tex].Standard deviation of 0.004 inch, hence [tex]\sigma = 0.004[/tex]Item a:
This probability is the p-value of Z when X = 0.753 subtracted by the p-value of Z when X = 0.75.
X = 0.753:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.753 - 0.753}{0.004}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5
X = 0.75:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a p-value of 0.2266
0.5 - 0.2266 = 0.2734
0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.
Item b:
This probability is the p-value of Z when X = 0.75 subtracted by the p-value of Z when X = 0.74, hence:
X = 0.75:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a p-value of 0.2266
X = 0.74:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]
[tex]Z = -3.25[/tex]
[tex]Z = -3.25[/tex] has a p-value of 0.0006.
0.2266 - 0.0006 = 0.226
0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.
Item c:
This probability is 1 subtracted by the p-value of Z when X = 0.76, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.76 - 0.753}{0.004}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599.
1 - 0.9599 = 0.0401
0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.
Item d:
This probability is the p-value of Z when X = 0.74, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]
[tex]Z = -3.25[/tex]
[tex]Z = -3.25[/tex] has a p-value of 0.0006.
0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.
A similar problem is given at https://brainly.com/question/24663213
Box has 10 M&M candies: 5 red and 5 blue.
Two candies are taken from this box.
Find the probability that the first randomly taken candy will be red and second will be red again.
Taken candy doesn't go back to the box.
Simplify final fraction.
Answer:
As a simplified fraction, the probability that the first randomly taken candy will be red and second will be red again is [tex]\frac{2}{9}[/tex]
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The order in which the candies are selected is not important. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Desired outcomes:
2 candies from a set of 5. So
[tex]D = C_{5,2} = \frac{5!}{2!(5-2)!} = 10[/tex]
Total outcomes:
2 candies from a set of 10. So
[tex]T = C_{10,2} = \frac{10!}{2!(10-2)!} = 45[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{10}{45} = \frac{2}{9}[/tex]
As a simplified fraction, the probability that the first randomly taken candy will be red and second will be red again is [tex]\frac{2}{9}[/tex]
An investment of $100 comma 000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned 8% interest, the second 6%, and the third 9%. Total interest from the investments was $ 7380. The interest from the first investment was 2 times the interest from the second. Find the amounts of the three parts of the investment.
Answer:
$54000 was invested at 8% interest$36000 was invested at 6% interest$10000 was invested at 9% interestStep-by-step explanation:
Total Investment = $100,000
The investment was split into three parts (say x, y and z)
x+y+z=100000Simple Interest = Principal X Rate/100 X Time
The first part of the investment earned 8% interest
Interest on the first part = 0.08x
The second part of the investment earned 6% interest
Interest on the second part = 0.06y
The third part of the investment earned 9% interest
Interest on the third part = 0.09z
Total interest from the investments = $ 7380.
Therefore:
0.08x+0.06y+0.09z=7380
The interest from the first investment was 2 times the interest from the second.
0.08x=2 X 0.06y
0.08x=0.12y
Substituting we have:
0.12y+0.06y+0.09z=7380
0.18y+0.09z=7380
From 0.08x=0.12y
x=1.5y
Substituting x=1.5y into x+y+z=100000
1.5y+y+z=100000
2.5y+z=100000
z=100000-2.5y
Substituting z=100000-2.5y into 0.18y+0.09z=7380
0.18y+0.09(100000-2.5y)=7380
0.18y+9000-0.225y=7380
-0.045y=-1620
Divide both sides by -0.045
y=36000
Recall: x=1.5y
x=1.5 X 36000
x =54000
x+y+z=100000
54000+36000+z=100000
z=100000-(54000+36000)
z=10,000
Therefore:
$54000 was invested at 8% interest$36000 was invested at 6% interest$10000 was invested at 9% interestIn a random sample of 25 laptop computers the mean repair cost was $150 with a standard deviation of $35. Compute the 99% confidence interval for u
Answer:
[tex]150-2.797\frac{35}{\sqrt{25}}=130.421[/tex]
[tex]150+2.797\frac{35}{\sqrt{25}}=169.579[/tex]
Step-by-step explanation:
Information given
[tex]\bar X=150[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean
s=35 represent the sample standard deviation
n=25 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=25-1=24[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex] and the critical value for this case [tex]t_{\alpha/2}=2.797[/tex]
Now we have everything in order to replace into formula (1):
[tex]150-2.797\frac{35}{\sqrt{25}}=130.421[/tex]
[tex]150+2.797\frac{35}{\sqrt{25}}=169.579[/tex]
how much of other chemicals must be evaporated from 400grams of a hand sanitizer that is 24% alcohol to strengthen it to a hand sanitizer that is 30% alcohol? correct your answer to the nearest whole number
Answer:
80 grams
Step-by-step explanation:
Weight of 24% solution = 400 grams
Alcohol content= 24%
Amount of alcohol= 400*0.24= 96 grams
Weight of 30% solution with same amount of alcohol:
96/0.3= 320 gramsThe difference in weights is the other chemicals evaporated:
400-320=80 gramsFind the population density of domesticated dogs if there are 43,600,000 dogs in the United States and the area of the United States is 3,794,083 square miles. Round to the nearest 10th
Answer:
11.49 dogs/ square mile
Step-by-step explanation:
Population density of any place is number of living being living in that area per unit area.
Population density = number of living being/ area of the place
given
number of dogs= 43,600,000
area = 3,794,083 square miles
population density of domesticated dogs = 43,600,000 /3,794,083
= 11.4915 dogs/ square mile
population density of domesticated dogs, rounded to the nearest 10th
is 11.49 dogs/ square mile.
Henry hoards a hundred hens. Every day, each hen lays an egg with probability 0.8 independently of all others. Henry sells each egg for 3 cents, except that on half the days, his dog Barkie breaks half the eggs laid that day (and 1.5 eggs is as sellable as 1), and on the remaining half of the days, Barkie breaks 30 eggs. What is the probability that Henry makes more than a $1.30 today
Answer:
99.27%
Step-by-step explanation:
We have to:
sample size (n) = 100
p = 0.8
1 egg = 3 * 0.01 = 0.03
eggs = 100 - 30 broken eggs = 70
Let's find the probability of winning more than $ 1.30 today:
p (x> 1.30) = p (z, (x -m) / sd)
average earnings:
per day would be:
m1 = 100 / 1.5 * 0.8 * 0.03
m1 = 1.6
sd1 = (m1 * (1 - p)) ^ (1/2) = (1.6 * (1 - 0.8)) ^ (1/2)
sd1 = 0.566
earnings for broken eggs:
m2 = 70 * 0.8 * 0.03
m2 = 1.68
sd2 = (m2 * (1 - p)) ^ (1/2) = (1.68 * (1 - 0.8)) ^ (1/2)
sd2 = 0.58
Now the total profit would be:
m = m1 + m2 = 1.6 + 1.68
m = 3.28
sd = sd1 ^ 2 + sd2 ^ 2 = 0.566 ^ 2 + 0.58 ^ 2
sd = 0.81
now yes, replacing:
p (x> 1.30) = p (z> (1.3 - 3.28) /0.81)
p (x> 1.30) = p (z> -2.44)
for this z = 0.0073
Therefore the probability would be:
1 - 0.0073 = 0.9927
That is, we would have a 99.27% probability of achieving the goal.
1. (a) The life time of a certain brand of bulbs produced by a company is normally distributed, with mean 210 hours and standard deviation 56 hours. What is the probability that a bulb picked at random from this company’s products will have a life time of:
(i) (ii) (iii) at least 300 hours,
at most 100 hours, between 150 and 250 hours.
(b) In a contest, two friends, Kofi and Mensah were asked to solve a problem. The probability that Kofi will solve it correctly is and the probability that Mensah
will solve it correctly is . Find the probability that neither of them solved it correctly.
2. Suppose that the random variable, X, is a number on the biased die and the p.d.f. of X is as shown below;
X
1
2
3
4
5
6
P(X=x)
1/6
1/6
1/5
k
1/5
1/6
a) Find;
(i) (ii) (iii) (iv) (v) the value of k. E(X)
E(X2) Var(X) P(1 £X <5)
b) If events A and B are such that they are independent, and P(A) = 0.3 with P(B) = 0.5;
i. ii. Find P(A n B) and P(AUB)
Are A and B mutually exclusive? Explain.
c) In how many ways can the letters of the word STATISTICS be arranged?
Answer:
See explanation
Step-by-step explanation:
Q1)a
- Denote a random variable ( X ) as the life time of a brand of bulb produced.
- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:
X ~ Norm ( 210 , 56^2 )
i) The bulb picked to have a life time of at least 300 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )
P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )
P ( X ≥ 300 ) = P ( Z ≥ 1.607 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054 .. Answer
ii) The bulb picked to have a life time of at most 100 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )
P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247 .. Answer
iii) The bulb picked to have a life time of between 150 and 250 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )
P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205 .. Answer
Q1)b
- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:
p ( A ) = 0.25
- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:
p ( B ) = 0.4
- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.
- So for independent events the required probability can be stated as:
p ( A' & B' ) = p ( A' ) * p ( B' )
p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]
p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]
p ( A' & B' ) = 0.45 ... Answer
Q2)a
- A discrete random variable X: defines the probability of getting each number on a biased die.
- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.
∑ p ( X = xi ) = 1
p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )
1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1
k = 0.1 ... Answer
- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:
E ( X ) = ∑ p ( X = xi ) . xi
E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6
E ( X ) = 3.5 .. Answer
- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:
E ( X^2 ) = ∑ p ( X = xi ) . xi^2
E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36
E ( X^2 ) = 15.233 .. Answer
- The variance of the discrete random distribution for the variable X can be determined from:
Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2
Var ( X ) = 15.2333 - [ 3.5 ] ^2
Var ( X ) = 2.9833 ... Answer
- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:
P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )
P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1
P ( 1 < X < 5 ) = 0.467 ... Answer
Q2)b
- Two independent events are defined by their probabilities as follows:
p ( A ) = 0.3 and p ( B ) = 0.5
- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.
- For the two events to occur simultaneously at the same time:
p ( A & B ) = p ( A )* p ( B )
p ( A & B ) = 0.3*0.5
p ( A & B ) = 0.15 ... Answer
- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:
p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )
p ( A U B ) = 0.3 + 0.5 - 2*0.15
p ( A U B ) = 0.5 ... Answer
- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:
p ( A & B ) = 0.15 ≠ 0
Q2)c
- The letters of the word given are to be arranged in number of different ways as follows:
STATISTICS
- Number of each letters:
S : 3
T : 3
A: 1
I: 2
C: 1
- 10 letters can be arranged in 10! ways.
- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:
[tex]\frac{10!}{3!3!2!} = \frac{3628800}{72} = 50,400[/tex]
- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.
Reese read twice as many pages Saturday than she read Sunday. If she read a total of 78 pages over the weekend, how many pages did Reese read Sunday?
Answer:
Reese read 26 pages on Sunday.
Step-by-step explanation:
If she read twice the amount on Saturday than she read on Sunday, then you can replace the days with variables. 2x for Saturday and x for Sunday. That leaves you with 2x + x = 78. This can be simplified to 3x = 78. We can solve this by dividing both sides by 3. This leaves you with x = 26. If x represents the pages read on Sunday then Reese read 26 pages on Sunday.
Please see picture of problem
Answer:
w = 9, l = 12
Step-by-step explanation:
If w is the width and l is the length, then:
l = 2w − 6
The area of the rectangle is width times length.
A = wl
Substituting and solving:
108 = w (2w − 6)
108 = 2w² − 6w
54 = w² − 3w
0 = w² − 3w − 54
0 = (w + 6) (w − 9)
w = -6 or 9
The width is 9, so the length is 12.
John is painting parallel lines in the parking lot to create parking spaces. The measure of angle A is 60°. What is the measure of angle B? A. 60° B. 90° C. 120° D. 180° E. any acute angle
Answer:60°
Step-by-step explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me....
Answer:
A.
60°
Hope its right.
Step-by-step explanation:
Select one: A. ∠T ≅ ∠F B. ∠T ≅ ∠D C. ∠J ≅ ∠F D. ∠J ≅ ∠D
Answer:
B
Step-by-step explanation:
Since you are given line RT and ND, then all you need to do by ASA is the make sure that the angles are the two endpoints are congruent. Since the problem already gave you R and N, then all thats left is to relate the other two endpoints, namely, T and D
What is the base of a triangle that has a height of 6 centimeters and an area of 18 centimeters? Use the formula h = StartFraction 2 A Over b EndFraction, where A represents the area of the triangle, h represents the height, and b represents the length of the base. One-third centimeter Two-thirds centimeters 3 Centimeters 6 Centimeters
Answer:
The base is 6cm
Step-by-step explanation:
Using [tex]\frac{1}{2}[/tex]bh=area, you just plug in 6 for h and 18 for area and solve. The formula should be 3b=18, which equals 6
The solution is Option D.
The length of the base of the triangle is B = 6 cm
What is a Triangle?A triangle is a plane figure or polygon with three sides and three angles.
A Triangle has three vertices and the sum of the interior angles add up to 180°
Let the Triangle be ΔABC , such that
∠A + ∠B + ∠C = 180°
The area of the triangle = ( 1/2 ) x Length x Base
For a right angle triangle
From the Pythagoras Theorem , The hypotenuse² = base² + height²
if a² + b² = c² , it is a right triangle
if a² + b² < c² , it is an obtuse triangle
if a² + b² > c² , it is an acute triangle
Given data ,
Let the area of the triangle be represented as A
Now , the value of A = 18 cm²
Let the base of the triangle be B
Let the height of the triangle be H = 6 cm
Now , area of the triangle = ( 1/2 ) x Length x Base
Substituting the values in the equation , we get
18 = ( 1/2 ) x B x 6
18 = 3B
Divide by 3 on both sides of the equation , we get
B = 6 cm
Hence , the base length of the triangle is B = 6 cm
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Write the algebraic expression that represents the following:
a. The difference between six times a number and two
b. Five less than the quotient of nine and a number
c. One half of the sum of six times a number and twenty-two
d. Nine less than twice the difference between a number and seven
e. Eleven less than a number squared
Question 1
A student said, "To find the value of 109.2 : 6 I can divide 1,092 by 60."
Do you agree with her? Explain your reasoning.
B 1 U
14
X,
E = = =
[D] TTT
12pt
Answer:
Yes, her reasoning is correct
Step-by-step explanation:
Given the ratio: 109.2 : 6
Written in fractional form, we have:
[tex]109.2: 6=\dfrac{109.2}{6}[/tex]
Now:
[tex]\dfrac{109.2}{6}=\dfrac{109.2}{6}X1, $ Let 1=\dfrac{10}{10}, \\\\=\dfrac{109.2}{6}X\dfrac{10}{10}\\\\=\dfrac{1092}{60}[/tex]
Therefore, the student's reasoning is correct. In fact, as a check:
[tex]\dfrac{109.2}{6}=18.2\\\\\dfrac{1092}{60}=18.2[/tex]
We would obtain the same result in both cases.
Answer:
yes
Step-by-step explanation:
5y−4≥26 first step?
Answer:
5y - 4 ≥ 26
5y ≥ 26 + 4
5y ≥ 30
y ≥ 30/5
y ≥ 6
Mark Brainliest .........
Answer:
work is shown and pictured
Todd is 3 years older than his brother Jack. If Jack is x years old and Todd is y years
old, write a rule that relates their ages over time. When Jack is 28 years old, how old will Todd be?
Answer:
1) y = x + 3
2) 31 years old
Step-by-step explanation:
Jack is x years old
Todd is y years old
Todd is 3 years older than his brother Jack => y = x + 3
When Jack is 28 years old, Todd is y = 28 + 3 = 31 years old
A parent needs up to twelve students for a game. He needs no fewer than five male students. Let x represent the number of female students and y represent the number of male students. Select all inequalities that model this situation. x≥0 y>5 x+y≤12 x+y 0
Answer:
y>5, x+y underlined<12
Step-by-step explanation:
Answer:
Step-by-step explanation:
Which of the following tables shows a valid probability density function? a. x P(X=x) 0 38 1 14 2 38 b. x P(X=x) 0 0.2 1 0.1 2 0.35 3 0.17 c. x P(X=x) 0 910 1 −310 2 310 3 110 d. x P(X=x) 0 0.06 1 0.01 2 0.07 3 0.86 e. x P(X=x) 0 12 1 18 2 14 3 18 f. x P(X=x) 0 110 1 110 2 310 3
Answer:
Step-by-step explanation:
Since we know that for a distribution be a probability density function sum of all the probability events should be equal to 1 and all individual events should have probability between 0 and 1
a. x P(X=x)
0 -----3/8
1 -----1/4
2 -----3/8
P(X=0)+P(X=1)+P(X=2) = 3/8 + 1/4 + 3/8
P(X=0)+P(X=1)+P(X=2) = 6/8 + 2/8 = 1
This is a probability density function
b. x P(X=x)
0 ----0.2
1 ----0.1
2 ----0.35
3 ----0.17
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.2 + 0.1 + 0.35 + 0.17
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.65 + 0.17 = 0.82 ≠ 1
Therefore this is NOT a probability density function
c. x P(X=x)
0---- 9/10
1 ---- −3/10
2 ---- 3/10
3 ---- 1/10
Since P(X=1) is not between 0 and 1
Therefore this is NOT a probability density function
d. x P(X=x)
0 ----0.06
1 ----0.01
2 ----0.07
3 ----0.86
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.06 + 0.01 + 0.07 + 0.86
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.14 + 0.86 = 1
Therefore this is a probability density function
e. x P(X=x)
0 ----1/2
1 ----1/8
2 ----1/4
3 ----1/8
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/8 + 1/4 + 1/8
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/2 + 1/2 = 1
Therefore this is a probability density function
f. x P(X=x)
0 ----1/10
1 ----1/10
2 ----3/10
3 ----1/5
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1/10 + 1/10 + 3/10 + 1/5
P(X=0)+P(X=1)+P(X=2)+P(X=3) = 2/10 + 5/10 = 7/10 ≠ 1
Therefore this is NOT a probability density function
When is a rectangle a square?
Answer:
A. When its sides are congruent
Step-by-step explanation:
By definition a square is a four sided polygon (shape) which has four right angles and four sides of equal length.
So, when a rectangle has all congruent sides, its a square because its a rectangle so it has four right angles, and congruent sides are sides of the same length.
B is incorrect because a rectangle can have four right angles but have differing side lengths
C is incorrect because again a rectangle can have parallel sides but have differing side lengths
D is incorrect because it can have convex angles but again have differing side lengths
When sides of rectangle are congruent, then it is a square. Therefore, the correct answer is option A.
A rectangle is a closed two-dimensional figure with four sides. The opposite sides of a rectangle are equal and parallel to each other and all the angles of a rectangle are equal to 90°.
A square is a quadrilateral with four equal sides. There are many objects around us that are in the shape of a square. Each square shape is identified by its equal sides and its interior angles that are equal to 90°.
If all the sides of rectangle are congruent, then it is square.
Therefore, the correct answer is option A.
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Consider Statement A and Statement B below:
1. If neither of two real numbers is zero, then their product is also not zero.
2. If a and b are two real numbers, and if ab = 0, then either a = 0 or b= 0.
A. These two statements are equivalent because statement A is the converse of statement B.
B. These two statements are equivalent because statement A is the contrapositive of statement B.
C. These two statements are not equivalent.
Answer:
These two statements are equivalent because statement A is the contrapositive of statement B.
Step-by-step explanation:
In logic, when we have an "if" statement we can have its contrapositive or its converse.
Given a "if p, then q" (p is the hypothesis and q is the conclusion), the converse is "if q, then p", in other words, we interchange the hypothesis and the conclusion.
Now, given a "if p, then q", the contrapositive is "if not q, then not p". In other words, we take the negation of both the hypothesis and the conclusion and then we interchange them.
Now, let's take a look at our statements:
If a and b are two real numbers, and if ab = 0, then either a = 0 or b = 0.
In this case:
p = a and b are two real numbers and ab=0
q = either a = 0 or b = 0
Now, let's take the negative of p and q:
The negative of p would be: a and b are two real numbers and their product is not zero.
The negative of q would be: neither of the two real numbers is zero
Now, given than the contrapositive is "if not q, then not p" we would have:
If neither of two real numbers is zero, then their product is not zero.
We can see that this last sentence is the contrapositive of the first one and thus:
These two statements are equivalent because statement A is the contrapositive of statement B.
