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3 years ago

C/8+25=20 what does c equal

Mathematics

2 answers:

Korolek [52]3 years ago

8 0

Answer:

-40

Step-by-step explanation:

ollegr [7]3 years ago

5 0

Answer:

C=-40

Step-by-step explanation:

C/8+25=20

C/8=-5

C=-40

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9. The results of a random survey is displayed in the histogram as shown.

natali 33 [55]

The number of people in the sample survey is 75.

Survey sampling, as used in statistics, is the process of choosing a sample of constituents from a target population to perform a survey. The word "survey" can be used to describe a wide range of observational methods or procedures. Most frequently, a questionnaire meant to assess people's traits and/or opinions is employed in survey sampling. The topic of survey data collecting involves various methods of contacting sample participants after they have been chosen. Sampling is done to cut down on the expense and/or labor required to survey the complete target population. A census is a survey that includes all members of the target population. A sample is a subset or segment of a population from which data will be drawn.

The sample survey has

5 people in 1-4 miles per week

10 people in 5-8 miles per week

15 people in 9-12 miles per week

20 people in 13-15 miles per week

25 people in 16-19 miles per week

The number of people in the sample survey is

5 + 10 + 15 + 20 + 25

=5(1 + 2 + 3 + 4 + 5)

=5(15)

=75

Hence, the number of people in the sample survey is 75.

Learn more about sampling survey here-

brainly.com/question/24564273

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7 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

3 years ago

(-4, -2); y = - 2x + 4

padilas [110]

Answer:

y = -2x - 10

Step-by-step explanation:

Slope intercept form of equation is of form

y = mx+c

where m is the slope of line and c is the y intercept of the line.

Y intercept is point on y axis where the line intersects the y axis.

_____________________________

Given equation

y = -2x +4

comparing it with y = mx+ c

m = -2 , c = 4

_____________________________

when two lines are parallel, their slopes are equal.

Let the equation of new line in slope intercept form be y = mx + c

Thus slope of of new required line is -2

Thus m for new line is -2.

now, equation of required line : y = -2x+c

Given that this line passes through (-4, -2). This point shall should satisfy equation y = -2x+c.

Substituting the value of (-4, -2) we have

-2 = -2(-4)+c

=> -2 = 8 +c

=> -2 -8 = c

=> c = -10.

Thus , equation of required line is y = -2x - 10.

8 0

3 years ago

An instructor wants to write a test with 25 questions where each question is worth 3, 4, or 5 points based on difficulty. He wan

irakobra [83]

ANSWER

Find out the how many 3, 4, and 5 point questions could there be.

To proof

Let us assume that the 3 points based question be = x

Let us assume that the 4 points based question be = y

Let us assume that the 5 points based question be = z

As given

An instructor wants to write a test with 25 questions

than the equation become in the form

x + y + z = 25

he quiz to be worth a total of 100 points.

than the equation is becomes

3x + 4y + 5z =100

As given

He wants the number of 3-point questions to be 4 less than the number of 4-point questions

x = y -4

Than the three equation are

x + y + z = 25 ,3x + 4y + 5z =100 and x = y -4

put x = y -4 in the x + y + z = 25 ,3x + 4y + 5z =100

than

2y + z = 29, 7y +5z= 112

multiply 2y + z = 29 by 5 and subtracted 7y +5z= 112

10 y -7y + 5y -5y = 145 -112

3y = 33

$y = \frac{33}{3}$

y =11

put in the x = y -4

x = 11-4

x= 7

put the value of x ,y in the x + y + z = 25

7 + 11 +z =25

18 +z = 25

z = 25 -18

z=7

therefore

numbers of the 3 point question be = 7

numbers of the 4 point question be = 11

numbers of the 5 point question be = 7

Hence proved

3 0

3 years ago

How many lengths of pipe 2 1/3 ft long can be cut from a pipe that is 63 ft long? Assume there is no kerf.

Zinaida [17]

The answer is 27ft long

7 0

2 years ago