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2 years ago

5

Each person in a group of students was identified by year and asked when he or she preferred taking classes: in the morning, aft

ernoon, or evening. The results are shown in the table. Find the probability that the student preferred morning classes given he or she is a freshman. Round your answer to the nearest thousandth.
Freshman Sophomore Junior Senior
Morning: 19 2 6 16
Afternoon: 17 3 13 15
Evening: 8 14 9 7
A. 0.105
B. 0.182
C. 0.386
D. 0.432

1 answer:

Helen [10]2 years ago

7 0

Answer: 0.432

Step-by-step explanation: I just took the test!

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Last year 80 students signed up for a summer trip to Washington, D.C. This summer 50 students have signed up to go. What is the

Fynjy0 [20]

Given:

Last year 80 students signed up for a summer trip to Washington, D.C.

This summer 50 students have signed up to go.

To find:

The percent decrease in the number of students.

Solution:

We have,

Students in last year = 80

Students in this year = 50

Now,

$\text{Decrease}\%=\dfrac{\text{Students in last year - Students in this year }}{\text{Students in last year }}\times 100$

$\text{Decrease}\%=\dfrac{80-50}{80}\times 100$

$\text{Decrease}\%=\dfrac{30}{80}\times 100$

$\text{Decrease}\%=\dfrac{3}{8}\times 100$

$\text{Decrease}\%=37.5$

Therefore, the correct option is B.

7 0

2 years ago

Will give brainliest answer!

kifflom [539]

5x + 60y = 35

x +y = 1.5 : rewrite as x = 1.5-y and substitute this formula for x in the first one:

5(1.5-y) + 60y = 35

distribute:

7.5 - 5y + 60y = 35

combine like terms:

7.5 + 55y = 35

subtract 7.5 from both sides:

55y = 27.5

divide both sides by 55 to solve for y

y = 27.5 / 55 = 0.5

now substiute 0.5 for y in the 2nd equation:

x + 0.5 = 1.5

x = 1.5 - 0.5 = 1

he walked for 1 hour

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3 years ago

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

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2 years ago

The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of vi

baherus [9]

Answer: 2185

Step-by-step explanation:

Let p be the proportion of visitors are campers.

Given : The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state.

The prior proportion of visitors are campers : p=0.35

Allowable error : E= 2%= 0.02

We know that the z-value for 95% confidence = $z_c=1.96$

Then by Central Limit Theorem , the required sample size would be :

$n=p(1-p)(\dfrac{z_{c}}{E})^2$

$\Rightarrow\ n=0.35(1-0.35)(\dfrac{1.96}{0.02})^2\\\\ n= 0.35(0.65)(9604)$

Simply , we get

$n=2184.91\approx2185$ [Rounded to the next whole number.]

Hence, the smallest sample size to estimate the population proportion of campers =2185

7 0

3 years ago

The circumference of a circle is 11 pi inches.

ser-zykov [4K]

9514 1404 393

Answer:

30.25π square inches

Step-by-step explanation:

You can use the formula for area in terms of circumference:

A = C²/(4π)

A = (11π)²/(4π) = (121/4)π = 30.25π . . . square inches

_____

You may be expected to find the radius first:

C = 2πr ⇒ r = C/(2π) = 11π/(2π) = 5.5 . . . inches

Then use the area formula:

A = πr² = π(5.5 in)² = 30.25π in²

8 0

3 years ago