Answer:
[tex]-4\pm 3\sqrt{3}[/tex]
Step-by-step explanation:
[tex]3(x+4)^2-81=0 \\\\3(x^2+8x+16)-81=0 \\\\3x^2+24x+48-81=0 \\\\3x^2+24x-33=0 \\\\x=\dfrac{-24\pm\sqrt{24^2+4(3)(33)}}{2(3)}= \\\\\dfrac{-24\pm\sqrt{972}}{6}= \\\\\dfrac{-24\pm 18\sqrt{3}}{6}= \\\\-4\pm 3\sqrt{3}[/tex]
Hope this helps!
What is the area of a circle with a radius of 6 inches?
9. in.
O 125 in.
O 36in 2
81 in 2
Answer:
A = 36pi in^2
Step-by-step explanation:
The area of a circle is given by
A = pi r^2
A = pi 6^2
A = 36pi in^2
Radius(r)=6 inches
Area of circle=?
Now,
Area of circle=pi r ^2
=pi (6)^2
36 pi inches^2
So the right answer is 36 pi inches^2
Hope it helps
Imagine we are throwing a five-sided die 50 times. On average, out of these 50 throws how many times would this five-sided die show an odd number (1, 3 or 5)?
Answer:
[tex]30[/tex]
Step-by-step explanation:
There are 3 outcomes out of 5 that are odd.
[tex]\frac{3}{5}[/tex]
Throw it 50 times.
[tex]50*\frac{3}{5}[/tex]
Answer:
30.
Step-by-step explanation:
The probability of throwing a 1, 3 or 5 on one throw is 3/5.
So on 50 throws the average would be 50 * 3/5 = 30 times.
The distribution of the scores on a certain exam is N(40,5), which means that the exam scores are Normally distributed with a mean of 40 and standard deviation of 5. a. Sketch the curve and label, on the x-axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 50. Shade the region under the Normal curve whose area corresponds to this probability.
Answer:
Step-by-step explanation:
Let x represent the random variable representing the scores in the exam. Given that the scores are normally distributed with a mean of 40 and a standard deviation of 5, the diagram representing the curve and the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations is shown in the attached photo
1 standard deviation = 5
2 standard deviations = 2 × 5 = 10
3 standard deviations = 3 × 5 = 15
1 standard deviation from the mean lies between (40 - 5) and (40 + 5)
2 standard deviations from the mean lies between (40 - 10) and (40 + 10)
3 standard deviations from the mean lies between (40 - 15) and (40 + 15)
b) We would apply the probability for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = sample mean
µ = population mean
σ = standard deviation
From the information given,
µ = 40
σ = 5
the probability that a randomly selected score will be greater than 50 is expressed as
P(x > 50) = 1 - P( ≤ x 50)
For x = 50,
z = (50 - 40)/5 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.98
P(x > 50) = 1 - 0.98 = 0.02
What is the quotient? Negative StartFraction 3 over 8 EndFraction divided by negative one-fourth WILL GIVE BRAINLIEST
Answer:
Im not too sure given yopur formatting, but I believe you meant this: -((3/8)/(-1/4))
therfore, the answer is: 3/2
Step-by-step explanation:
You can seperate the two fractions as 3/8 divided by -1/4, which is equal to 3/8 * 4/-1, simply multiply across to get 12/-8, which simplifies to -3/2. Flip the sign to get 3/2
Answer:
3/2
Step-by-step explanation:
I will give brainliest if you answer this question !!!
Answer:
A) 2
Step-by-step explanation:
Start off with the given information. The question states that the x-int. is 4, so you should recognize that there is a point at (4,0). Plug the point into the equation.
k(4) + 2(0) + 8 = 0
Now simplify the equation.
4k + 0 + 8 = 0
Isolate the variable, make sure it's on its own side.
4k = 8
Now get the k by itself to solve the equation. Divide both sides by 4.
k = 2
Si: P(x) = 3x ; R(x) = 2x ; calcula : P(R (0))
Answer:
0
Step-by-step explanation:
Put 0 where x is and do the arithmetic.
R(0) = 2(0) = 0
P(R(0)) = P(0) = 3(0)
P(R(0)) = 0
Suppose a random variable, x, follows a Poisson distribution. Let μ = 2.5 every minute, find the P(X ≥ 125) over an hour. Round answer to 4 decimal places.
Answer:
P(X ≥ 125) = 0.9812
Step-by-step explanation:
To solve this question, we need to understand the Poisson distribution and the normal distribution.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\lambda[/tex] is the mean in the given interval, which is the same as the variance.
Normal distribution:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The Poisson can be approximated to the normal, with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex]
Let μ = 2.5 every minute
This is the mean of the Poisson, so [tex]\lambda = 2.5n[/tex], in which n is the number of minutes.
P(X ≥ 125) over an hour
An hour has 60 minutes, so [tex]n = 60, \lambda = 2.5*60 = 150, \sigma = \sqrt{150} = 12.25[/tex]
Using continuity correction, this is [tex]P(X \geq 125 - 0.5) = P(X \geq 124.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 124.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{124.5 - 150}{12.25}[/tex]
[tex]Z = -2.08[/tex]
[tex]Z = -2.08[/tex] has a pvalue of 0.0188
1 - 0.0188 = 0.9812
So
P(X ≥ 125) = 0.9812
A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform the required hypothesis test at the given significance level. Use the critical -value approach.
= 20.5, n = 11 , σ = 7, H0: μ = 18.7; Ha: μ ≠ 18.7, α = 0.01
Answer:
Z = 0.8528 < 2.576
The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance
Null hypothesis is Accepted at 0.01 level of significance.
There is no significance difference between the means
Step-by-step explanation:
Given data
size of the sample 'n' = 11
mean of the sample x⁻ =20.5
Mean of the Population μ = 18.7
Standard deviation of Population σ = 7
Test statistic
[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]Z = \frac{20.5 -18.7}{\frac{7}{\sqrt{11} } }[/tex]
[tex]Z = \frac{1.8}{2.1105}[/tex]
Z = 0.8528
critical Value
[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} = 2.576[/tex]
The calculated value Z = 0.8528 < 2.576 at 0.01 level of significance
Null hypothesis is Accepted at 0.01 level of significance.
There is no significance difference between the means
Answer:Answer:
B. 18.7 ± 9.7
Step-by-step explanation:
Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of cigarettes. Construct a 99% confidence interval for the mean nicotine content of this brand of cigarette.
a 27.3 ± 3.033
b 27.3 ± 3.321
c 27.3 ± 3.217
d 27.3 ± 3.131
Part II - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%. Do you agree?
a No, since the value 28.4 does not fall in the 98% confidence interval.
b
Yes, since the value 28.4 does fall in the 98% confidence interval.
c
Yes, since the value 28.4 does not fall in the 98% confidence interval.
d
No, since the value 28.4 does fall in the 98% confidence interval.
Answer:
(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].
(II) No, since the value 28.4 does not fall in the 98% confidence interval.
Step-by-step explanation:
We are given that a new cigarette has recently been marketed.
The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.
Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean nicotine content = 27.3 milligrams
s = sample standard deviation = 2.8 milligrams
n = sample of cigarettes = 9
[tex]\mu[/tex] = true mean nicotine content
Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
Part I : So, 99% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-3.355 < [tex]t_8[/tex] < 3.355) = 0.99 {As the critical value of t at 8 degree
of freedom are -3.355 & 3.355 with P = 0.5%}
P(-3.355 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 3.355) = 0.99
P( [tex]-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
P( [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.99
99% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-3.355 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+3.355 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] , [tex]27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }[/tex] ]
= [27.3 [tex]\pm[/tex] 3.131]
= [24.169 mg , 30.431 mg]
Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].
Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.
The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.
The Pivotal quantity for 98% confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean nicotine content = 24.9 milligrams
s = sample standard deviation = 2.6 milligrams
n = sample of cigarettes = 9
[tex]\mu[/tex] = true mean nicotine content
Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.
So, 98% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.896 < [tex]t_8[/tex] < 2.896) = 0.98 {As the critical value of t at 8 degree
of freedom are -2.896 & 2.896 with P = 1%}
P(-2.896 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.896) = 0.98
P( [tex]-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
P( [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.98
98% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.896 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.896 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] , [tex]24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }[/tex] ]
= [22.4 mg , 27.4 mg]
Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].
No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.
Rectangular pieces of land 50m wide and 120m long partitioned to two parts along diagonal what is the length meter of the fence covering one part of it?
Answer:
a
Step-by-step explanation:
bc I just took the quiz☝
True or False? The segments shown below could form a triangle.
Answer:
True
Step-by-step explanation:
The rule is that every 2 sides' sum is greater than the other side.
So, let's test it.
9+4>11
9+11>4
11+4>9
it checks out!
What similarity statement can you write relating the three triangles in the diagram? Select one: a. ∆WUT ∼ ∆VUW ∼ ∆WVT b. ∆TWV ∼ ∆VUW ∼ ∆UWT c. ∆UVW ∼ ∆WUT ∼ ∆WVT d. ∆UVW ∼ ∆UWT ∼ ∆WVT
The correct option is d. ∆UVW ∼ ∆UWT ∼ ∆WVT
The following information should be considered:
The altitude of the hypotenuse of the right traingle splits the triangle into 2 triangles But they should be the same to the real triangle.learn more: https://brainly.com/question/994316?referrer=searchResults
Answer:
UVW ~ ΔUWT ~ ΔWVT
Step-by-step explanation:
Mr. Collins and Ms. LaPointe are saving money to buy a new motorcycle. The total amount of money Mr. Collins will save is given by the function p(x) = 62 + 5x. The total amount of money Ms. LaPointe will save is given by the function a(x) = x2 + 38. After how many months, x, will they have the same amount of money saved?
We have been given that Mr. Collins and Ms. LaPointe are saving money to buy a new motorcycle. The total amount of money Mr. Collins will save is given by the function [tex]p(x) = 62+5x[/tex]. The total amount of money Ms. LaPointe will save is given by the function [tex]a(x) = x^2+38[/tex]. We are asked to find number of months when they have the same amount of money saved.
To solve our given problem, we will equate both equations as:
[tex]x^2+38=62+5x[/tex]
[tex]x^2-5x+38=62+5x-5x[/tex]
[tex]x^2-5x+38=62[/tex]
[tex]x^2-5x+38-62=62-62[/tex]
[tex]x^2-5x-24=0[/tex]
[tex]x^2-8x+3x-24=0[/tex]
[tex]x(x-8)+3(x-8)=0[/tex]
[tex](x-8)(x+3)=0[/tex]
[tex](x-8)=0,(x+3)=0[/tex]
[tex]x=8,x=-3[/tex]
Since time cannot be negative, therefore, after 8 months they will have the same amount of money saved.
00:00
Use the drop-down menu to complete the comparison
1 mi 72 in
Choose...
5,286 ft
Answer:
5286 ft
Step-by-step explanation:
The question has been answered but I guess you need an explanation on how the answer was derived.
Given Parameter
1 mi 72 inch
Required
Convert to feet
To complete this conversion, we need to split the given data into 2.
1 mi and 72 inch
= 1 mi + 72 inch
Then we convert them separately.
From unit of standard conversion;
1 mile (1 mi) is equivalent to 5280 ft
Also from unit of standard conversion
1 inch = 1/12 foot
So, 72 inches will be
=> 72 * 1/12 ft
=> 72/12 ft
=> 6 ft
Now that we have the equivalent of 1 mi and 72 in, in foot.
We can then write the following;
1 mi + 72 inch
=> 5280 ft + 6 ft
=> 5286 ft.
Hence, 1 mi 72 inch is equivalent to 5286 ft
Answer:
Use the drop-down menu to complete the comparison.
1 mi 72 in.
=
5,286 ft
Step-by-step explanation:
Lines a and b are parallel.
Parallel lines a and b are cut by transversals s and t. The angles formed by the intersection of lines a, s, and t, clockwise from top left are (7 x) degrees, blank; formed by lines b and s are blank, blank, blank, (11 x) degrees; formed by lines b and t are 90 degrees,
What is the value of x?
5
10
35
55
If you answer correctly, we will name the brainiest!
This is a challenge question, you have 5 minutes.
Answer:
The answer is x=5.
Step-by-step explanation:
11x + 7x + 90 = 180
18x + 90 = 180
-90 -90
= 18x = 90
x = 5
x is 5!
Value of x is 10°.
Correct option is B.
What is angle?A figure which is formed by two rays or lines that shares a common endpoint is called an angle. The word “angle” is derived from the Latin word “angulus”, which means “corner”.
Given,
a || b
cut by transversal s and t
lines b and t have angle = 90°
The first top left angle between a, s, t = 7x
The last bottom angle between s and b = 11x
a || b, so a and t will also have angle = 90°
by the figure,
180° - 11x + 90° - 7x + 90° = 180°
360° - 18x = 180°
18x = 180°
x = 10°
Hence, 10° is value of x.
Learn more about angle here:
brainly.com/question/28451077
#SPJ6
Which function represents exponential decay? f(x) = f(x) = f(x) = 4 f(x) = 4
Answer:
f(x)=4
Step-by-step explanation:
In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time. It can be expressed by the formula y=a(1-b)x wherein y is the final amount, a is the original amount, b is the decay factor, and x is the amount of time that has passed.
f(x)=4(2/3)*
option D
opción D
a game board has 9 cards, and 3 say WIN. Emma picks 2 cards without replacing the first. What is the probability that neither say WIN.
Answer:
0.417
Step-by-step explanation:
Total event = 9
3 cards are for win
6 cards are for not win .ie neither win
He picks two cards without replacement.
Probabilty of neither to say win
= 6/9 * 5/8
=30/72
= 0.417
Water makes about 3/4 of a person’s weight. If a student weighs 30 lbs., how much of his weight is water?
Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 32 mpg with a standard deviation of 6 mpg. Thirty-one non-hybrid sedans get a mean of 21 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test at the 5% level to evaluate the manufacturers claim. NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
Answer:
There is enough evidence to support the claim of the manufacturers that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
Step-by-step explanation:
This is a hypothesis test for the difference between populations means.
The claim is that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0[/tex]
The significance level is 0.05.
The sample 1 (hybrid), of size n1=21 has a mean of 32 and a standard deviation of 6.
The sample 2 (non-hybrid), of size n2=31 has a mean of 21 and a standard deviation of 3.
The difference between sample means is Md=11.
[tex]M_d=M_1-M_2=32-21=11[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{6^2}{21}+\dfrac{3^2}{31}}\\\\\\s_{M_d}=\sqrt{1.714+0.29}=\sqrt{2.005}=1.4158[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{11-0}{1.4158}=\dfrac{11}{1.4158}=7.77[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-2=21+31-2=50[/tex]
This test is a right-tailed test, with 50 degrees of freedom and t=7.77, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=P(t>7.77)=0.0000000002[/tex]
As the P-value is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones.
A population of values has a normal distribution with μ = 247 and σ = 62.2. You intend to draw a random sample of size n = 16. (a) Find the probability that a single randomly selected value is greater than 295.2. (b) Find the probability that a sample of size n= 16 is randomly selected with a mean greater than 295.2. Give your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer:
a) [tex] z = \frac{295.2-247}{62.2}=0.772[/tex]
And using the normal distribution table or excel we got:
[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]
b) [tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]
And we can use the normal standard table or excel in order to find the probability and we got:
[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]
Step-by-step explanation:
For this case we know that the random variable of interest is normally distributed with the following parameters:
[tex] X \sim N (\mu = 247, \sigma =62.2)[/tex]
Part a
We want to find this probability:
[tex] P(X>295.2)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{X-\mu}{\sigma}[/tex]
Replacing we got:
[tex] z = \frac{295.2-247}{62.2}=0.772[/tex]
And using the normal distribution table or excel we got:
[tex] P(Z>0.772) = 1-P(Z<0.772) = 1-0.77994= 0.22006[/tex]
Part b
We select a random sample of size n = 16 and we try to find this probability:
[tex] P(\bar X >295.2)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z = \frac{295.5 -247}{\frac{62.2}{\sqrt{16}}}= 3.119[/tex]
And we can use the normal standard table or excel in order to find the probability and we got:
[tex] P(z>3.119) =1-P(Z<3.119)= 1- 0.999093=0.000907[/tex]
Suppose we have collected times, in minutes, that it takes volunteers to complete a set of pencil and paper mazes. Volunteers are randomly assigned to one of two groups. Group 1 watch a 5 minute video explaining good strategies for completing the mazes. Group 2 watch a 5 minute video of other people successfully completing the mazes, but with no explanation given.
Researchers are interested in testing against the null hypothesis that there is no difference in population mean times that it takes people to complete mazes after watching either type of video.
There are the summary statistics after collecting data. Note that the sample sizes differ, because some subjects did not follow instructions correctly and so they were not included in the analysis.
x1 = 11.98, x2 = 9.15
s1 = 8.69, s2 = 7.75
n1 = 43, n2 = 52
a. Is this an experimental or observational study?
Observational
Experimental
b. What are the predictor and response variables, and are they categorical or quantitative? (You must get all correct)
Video type is categorical
Finishing time is quantitative
Video type is the response variable
Video type is the predictor variable
Video type is quantitative
Finishing time is categorical
Finishing time is the predictor variable
Finishing time is the response variable
How should the null hypothesis be written?
H0: x1 - x2 = 0
H0: μ1 - μ2 = 5
H0: μ = 0
H0: x = 5
H0: x1 - x2 = 5
H0: μ1 - μ2 = 0
(for the following calculations, round to 2 decimal places)
c. Calculate the standard error of x1 - x2:
d. Calculate the approximate 95% CI for μ1 - μ2: to
e. The result of the hypothesis test is:
We do not have evidence that there is a difference in population means, because the null value is inside the 95% CI.
We do not have evidence that there is a difference in population means, because the null value is outside the 95% CI.
We have evidence that there is a difference in population means, because the null value is inside the 95% CI.
We have evidence that there is a difference in population means, because the null value is outside the 95% CI.
f. What kind of error is it *possible* that we have made?
1. It is possible that we made a Type I error, because this is when you reject a false null hypothesis.
2. It is possible that we made a Type I error, because this is when you fail to reject a false null hypothesis.
3. It is possible that we made a Type I error, because this is when you reject a true null hypothesis.
4. It is possible that we made a Type I error, because this is when you fail to reject a true null hypothesis.
5. It is possible that we made a Type II error, because this is when you reject a false null hypothesis.
6. It is possible that we made a Type II error, because this is when you fail to reject a false null hypothesis.
7. It is possible that we made a Type II error, because this is when you reject a true null hypothesis.
8. It is possible that we made a Type II error, because this is when you fail to reject a true null hypothesis.
Answer:
Step-by-step explanation:
Hello!
A group of volunteers is randomly assigned to two groups:
Group 1: watch a 5-minute video explaining good strategies for completing the mazes.
X[bar]₁ = 11.98, S₁ = 8.69, n₁ = 43
Group 2: watch a 5-minute video of other people successfully completing the mazes, but with no explanation given.
X[bar]₂ = 9.15, S₂ = 7.75, n₂ = 52
After watching the videos the volunteers were given a set of pencil and paper mazes to resolve and the time, in minutes, it took them to resolve the mazes was measured.
a)
An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information. These types of studies are usually made as precursors to a more formal experimental study, to have an idea of what's to be expected from the population.
An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It is also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).
⇒ Considering these definitions, and the description of the experiment, how the volunteers were treated exactly the same except for the videos and that the assignment of the groups was random, you can classify it as an experimental study.
b.
The response variable is the one that was measured by the researchers.
X: Time it takes the volunteer to complete the paper mazes.
This variable is quantitative continuous.
The predictor variable is the variable suspected to modify the response variable:
Y: Type of video assigned to the volunteer. Categorized: "Video on good strategies to solve mazes" and "Video showing people solving mazes"
This variable is a qualitative categorical.
If you want to compare the times it takes the volunteers of both groups the best is to do so trough the population means, so the parameter of interest is: μ₁ - μ₂
The claim is that there is no difference between the times that it takes people to complete mazes after watching either video so the hypotheses are:
H₀: μ₁ - μ₂=0
H₁: μ₁ - μ₂≠0
c.
Assuming X[bar]₁≈N and X[bar]₂≈N (since both samples n₁ and n₂ are large enough you can approximate the distribution of the sample means using the central limit theorem)
(X[bar]₁-X[bar]₂)≈N(μ₁-μ₂;σ₁²/n₁+σ₂²/n₂)
The estimation of the variance σ₁²/n₁+σ₂²/n₂ is:
V(X)= [tex]\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} = \frac{(8.69)^2}{43} +\frac{(7.75)^2}{52}= 2.91[/tex]
Standard error= √V(X)= √2.91= 1.706
d.
(X[bar]₁-X[bar]₂)±[tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}}[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.96[/tex]
(11.98-9.15)±1.96*1.706
[-0.51; 6.17]minutes
e.
The 95% confidence interval contains the zero, so using this CI and at a complementary significance level of 5%, the test is not significant, which means that there is no evidence to reject the null hypothesis.
Correct option: We do not have evidence that there is a difference in population means, because the null value is inside the 95% CI.
f.
You can make two types of errors when deciding over a hypothesis test:
Type I error: Reject the null hypothesis when the hypothesis is true.
Type II error: Fail to reject the null hypothesis when the hypothesis is false.
Since the null hypothesis wasn't rejected, there is a chance that a type II error was committed.
The correct option is:
6. It is possible that we made a Type II error because this is when you fail to reject a false null hypothesis.
I hope this helps!
The human average gestation time is 266 days from conception. A researcher suspects that proper nutrition plays an important role and that poor women with inadequate food intake should have shorter gestation times even when given vitamin supplements. A random sample of 20 poor women given vitamin supplements throughout the pregnancy has mean gestation time from conception x = 256 days. Human gestation times are approximately Normal with standard deviation ? = 16 days. The P -value for the researcher
Answer:
[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex]
The p value for this case can be calculated with this probability:
[tex]p_v =P(z<-2.795)=0.0026[/tex]
Step-by-step explanation:
Information provided
[tex]\bar X=256[/tex] represent the sample mean
[tex]\sigma=16[/tex] represent the population standard deviation
[tex]n=20[/tex] sample size
[tex]\mu_o =266[/tex] represent the value that we want to test
z would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to verify
We need to conduct a hypothesis in order to see if inadequate food intake should have shorter gestation times, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 266[/tex]
Alternative hypothesis:[tex]\mu < 266[/tex]
The statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]z=\frac{256-266}{\frac{16}{\sqrt{20}}}=-2.795[/tex]
The p value for this case can be calculated with this probability:
[tex]p_v =P(z<-2.795)=0.0026[/tex]
If the ratio of similarity between two similar polygons is 4:3.2, this ratio converts to
1:8.
8:1.
4:5.
5:4.
Answer:
5 : 4
Step-by-step explanation:
4: 3.2
Multiply by 10 to get rid of the decimal
4*10 : 3.2* 10
40 : 32
Divide each side by 8 to simplify
40/8 : 32/ 8
5 : 4
Please answer this correctly I want genius or expert people to answer this correctly.
Answer:
18.84mm^2
Step-by-step explanation:
Surface area of a cylinder can be found with the following formula:
2πrh+2πr^2
r is radius
h is height
Plug our values in
2π1(2)+2π(1)^2
2π(2)+2π(1)
4π+2π
6π
6(3.14)
18.84
A company had a gross profit of $300,000 based on sales of $400,000. Its cost of goods sold equals $700,000 is this true or false?
Answer:
false
Step-by-step explanation:
Answer:
Its true
Step-by-step explanation:
I took the quiz for Intro to Careers in Finance
Question 10
A bank loaned out $29,000, part of it at the rate of 13% annual interest, and the rest at 4% annual interest.
The total interest earned for both loans was $2,195.00. How much was loaned at each rate?
Answer:
$11,500 was invested at 13%.
$17,500 was invested at 4%
Step-by-step explanation:
This is a simple interest problem.
The simple interest formula is given by:
[tex]E = P*I*t[/tex]
In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.
After t years, the total amount of money is:
[tex]T = E + P[/tex]
In this question:
Loans totaling 29,000.
P was invested at 13%
29000 - P was invested at 4%.
First investment:
Principal P.
Interest 13% = 0.13.
One year, so t = 1.
So
[tex]E_{1} = P*0.13*1[/tex]
[tex]E_{1} = 0.13P[/tex]
Second investment:
Principal 29000 - P.
Interest 4% = 0.04.
One year, so t = 1.
So
[tex]E_{2} = (29000-P)*0.04[/tex]
The total interest earned for both loans was $2,195.00.
This means that [tex]E_{1} + E_{2} = 2195[/tex]
So
[tex]E_{2} = 2195 - E_{1}[/tex]
So we solve the following system:
[tex]E_{1} = 0.13P[/tex]
[tex]E_{2} = (29000-P)*0.04[/tex]
[tex]2195 - E_{1} = (29000-P)*0.04[/tex]
[tex]2195 - 0.13P = 1160 - 0.04P[/tex]
[tex]0.09P = 2195 - 1160[/tex]
[tex]P = \frac{2195 - 1160}{0.09}[/tex]
[tex]P = 11500[/tex]
$11,500 was invested at 13%.
29000 - 11500 = 17500
$17,500 was invested at 4%
Can someone help :(!
Answer:
1/(t+4)²
Step-by-step explanation:
[tex]\dfrac{t+3}{t+4}\div(t^2+7t+12)=\dfrac{t+3}{t+4}\cdot\dfrac{1}{(t+3)(t+4)}=\dfrac{(t+3)}{(t+3)(t+4)^2}\\\\=\boxed{\dfrac{1}{(t+4)^2}}[/tex]
The price of oil, in dollars per barrel, declined last week by 3.5%. If it started the week at 102.00 per barrel, at what per barrel price did it end the week?
Answer:
98.57
Step-by-step explanation:
3.5% over 100% X 102=3.43
102-3.43= 98.57
Tom drives a truck. His regular trip is a distance of 280 km. He drives at an average speed of 80 km/h.
For safety reasons Tom’s boss puts a speed limiter on his truck. This reduces Tom’s average speed by 10 km/h.
How much longer will it take Tom to drive his regular trip?
Answer:
.5 hour, or 30 minutes
Step-by-step explanation:
280 km / 80 kmh = 3.5 hr before limiter
280 km / (80-10) kmh = 4 hr after
so .5 hour from 3.5 to 4 hours. or 30 minutes
This is provided he matains this speed at all times.
What is the best estimate of the measure of the angle shown?
Answer:
141
I got it right
