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.
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]
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:
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
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]
What is the best estimate of the measure of the angle shown?
Answer:
141
I got it right
A sample of 1500 computer chips revealed that 41% of the chips fail in the first 1000 hours of their use. The company's promotional literature claimed that less than 44% fail in the first 1000 hours of their use. Is there sufficient evidence at the 0.02 level to support the company's claim?
Answer:
Yes, there will be sufficient evidence at the 0.02 level to support the company's claim
Step-by-step explanation:
Firstly we need to state the null and alternative hypothesis for the above question
We also need to test whether the proportion of chips that fails before 1000 hours is under 0.44. Hence, our hypotheses will be:
H o: p = 0.48
H a: p > 0.48
Where P represent the proportion of Chips.
the ortiz family is building a barn on a property with a perimeter of 120 meters. what is the area of the property that Mr. Ortiz is building the barn on?
Answer:
900 square meters
Step-by-step explanation:
The property is in the shape of a square.
The perimeter of a square is given as:
P = 4L
where L = length of its side
Therefore, the length of the property is:
120 = 4 * L
L = 120 / 4 = 30 meters
The area of a square is given as:
A = L * L
Therefore, the area of the property is:
A = 30 * 30 = 900 square meters
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!
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
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:
Solve. -2 1/3 - (-5)=
Answer:
[tex]\frac{8}{3}[/tex]
Step-by-step explanation:
[tex]-2\frac{1}{3} - (-5)\\=\frac{-7}{3} + 5\\= \frac{-7}{3} + \frac{15}{3} \\= \frac{8}{3}[/tex]
Hope this helps!
Answer:
8/3 or 2 2/3
Step-by-step explanation:
-2 1/3 - (-5)
-7/3 + 5
-7/3 + 15/3
8/3
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:
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☝
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.
An article in USA Today stated that Internal surveys paid for by directory assistance providers show that even the most accurate companies give out wrong numbers 15% of the time. Assume that you are testing such a provider by making 10 requests and also assume that the provider gives the wrong telephone number 15% of the time.
Required:
a. Find the probability of getting one wrong number.
b. Find the probability of getting at most one wrong number.
c. If you do get at most one wrong number, does it appear that the rate of wrong numbers is not 15%, as claimed?
Answer:
a. 11.26 % b. 6.76 %. It appears so since 6.76 % ≠ 15 %
Step-by-step explanation:
a. This is a binomial probability.
Let q = probability of giving out wrong number = 15 % = 0.15
p = probability of not giving out wrong number = 1 - q = 1 - 0.15 = 0.75
For a binomial probability, P(x) = ⁿCₓqˣpⁿ⁻ˣ. With n = 10 and x = 1, the probability of getting a number wrong P(x = 1) = ¹⁰C₁q¹p¹⁰⁻¹
= 10(0.15)(0.75)⁹
= 1.5(0.0751)
= 0.1126
= 11.26 %
b. At most one wrong is P(x ≤ 1) = P(0) + P(1)
= ¹⁰C₀q⁰p¹⁰⁻⁰ + ¹⁰C₁q¹p¹⁰⁻¹
= 1 × 1 × (0.75)¹⁰ + 10(0.15)(0.75)⁹
= 0.0563 + 0.01126
= 0.06756
= 6.756 %
≅ 6.76 %
Since the probability of at most one wrong number i got P(x ≤ 1) = 6.76 % ≠ 15 % the original probability of at most one are not equal, it thus appears that the original probability of 15 % is wrong.
To complete the square for x2 + 18x = 4, add what
to both sides
Answer:
subtract 4 to both sides ending with
x^2+18-4=0
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
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
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
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:
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%
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!
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.
