Solve for x. 5x+4=9 Can you help me out?

What is the height of the building? 33.9 m 37.1 m 36.8 m 34.5 m

madreJ [45]

37.1
Hope this helps a lot!!

5 0

3 years ago

A triangle is dilated by a scale factor of 1/5 and then rotated 90 degree clockwise about the origin. Why is the image similar t

yan [13]

Answers:
The corresponding angles of the triangles are congruent
The image is a reduction of the pre-image
Neither the dilation nor the rotation change the shape of the triangle

Explanation:
For shapes to be similar:
1- there should be a ratio between the sides
2- angles in first shape should be congruent to angles in second shape

Now, a scale factor of 0.2 means that the sides of the image are 0.2 of the length of the original shape. However, angles are not changes

Let's check the choices:
1- The corresponding sides of the triangles are congruent:
This option is incorrect as dilation changes the lengths of the sides

2- The corresponding angles of the triangles are congruent:
This option is correct as neither dilation nor rotation alters the measures of the angles

3- The corresponding sides of the image are 5 times as long as those of the pre-image:
This option is incorrect as the sides of the image are only 0.2 times as long as those of the pre-image

4- The image is a reduction of the pre-image:
This option is correct as the sides of the image are 0.2 times those of the pre-image which means that the shape is reduced

5- Neither the dilation nor the rotation change the shape of the triangle:
This option is correct as both dilation and rotation are rigid transformations that do not alter the shape of the triangle (a triangle remains a triangle only with different side lengths)

6- The rotation reduces the size of the triangle:
This option is incorrect as rotation does not alter the size of the shape. It only changes its position

Hope this helps :)

3 0

3 years ago

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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

3 years ago

Liling is a salesperson at a computer store she recurves a base salary of $2,300 per month and she receives a commission of 3% o

Harman [31]

Answer:

$3948

Step-by-step explanation:

Total earning:

Salary + commission + extended warrant fee
2300 + 0.03*44000 + 41*8 =
2300 + 1320 + 328 =
3948

8 0

2 years ago

there are 18 boys and 12 girl in math class. what is the ratio of girls to total students. A. 18:30 B. 12:30 C. 30:12 D. 30:18 2

zmey [24]

Answer:

1. B

2. B and C

3.C

4.B the answer is 238 but i guess the rounded

Step-by-step explanation:

3 0

3 years ago

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