Ask question

Ask question

All categories

2 years ago

13

What is the area of the trapezoid? A trapezoid has a base of 20 feet, a height of 24 feet, and a top side length of 16 feet. 368

ft2 432 ft2 480 ft2 528 ft2

2 answers:

asambeis [7]2 years ago

8 0

Answer:

432

Step-by-step explanation:

edgen

sesenic [268]2 years ago

5 0

Answer: the area of the trapezoid is 432 ft²

Step-by-step explanation:

The formula for determining the area of cross section of a trapezoid is expressed as

Area = 1/2(a + b)h

Where

a and b are the length of The bases are the 2 sides of the trapezoid which are parallel with one another.

h represents the height of the trapezoid.

From the information given,

a = 16 feet

b = 20 feet

h = 24 feet

Therefore,

Area of trapezoid = 1/2(16 + 20)24

= 1/2 × 36 × 24

= 432feet²

You might be interested in

Use the distributive property to rewrite the expressions. <br> 1. 4(3x-5) <br> 2. a(11+2b)

Nadusha1986 [10]

Answer:

1. 12x - 20

2. 11a + 2ab

4 0

2 years ago

Read 2 more answers

What is constant of proportionality ? y=14x k=?

Delvig [45]

Answer: I believe that its 14

Step-by-step explanation:

6 0

2 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

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

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0

2 years ago

The manager of a factory wants to compare the mean number of units assembled per employee in a week for two new assembly techniq

antoniya [11.8K]

Answer:

The most appropriate inferential statistic test in the present situation is the;

b. Two-sample t-test

Step-by-step explanation:

From the question, we have;

The number of subjects in each sample, n₁ = 100

The number of samples = 2

The nature of the test = To compare the two means

The value of the population standard deviation, σ = Unknown

From the given parameters, we have;

The t-test is a method of inferential statistics for determining whether the difference between the mean of two groups is significant, that is if the observed difference are due to chance, where the groups are likely to have some characteristics that are related

The t-test is utilized when testing an hypothesis, and it is used for small samples, where the population standard deviation is unknown

The types of t-test used are;

a) One-sample t-test

b) Two-sample t-test

c) Paired Sample t-test

Therefore, a two-sample t-test is most appropriate.

3 0

1 year ago

Joey puts 25% of his allowance into savings each week. If last week he put 10$ in savings, what was his allowance last week?

Yuri [45]

Answer:

40

Step-by-step explanation:

10=25% so multiply 10 by 4 and cause there is 4 25% in 100

7 0

2 years ago