Solve for the roots in the following equation. hint: factor both quadratic expressions. (x 4 5x 2 - 36)(2x 2 9x

A new car is purchased for 15300 dollars. The value of the car depreciates at 14.25% per year. What will the value of the car be

Andrei [34K]

Answer:

$\$6,082.70$

Step-by-step explanation:

we know that

The formula to calculate the depreciated value is equal to

$V=P(1-r)^{x}$

where

V is the depreciated value

P is the original value

r is the rate of depreciation in decimal

x is Number of Time Periods

in this problem we have

$P=\$15,300\\r=14.25\%=0.1425\\x=6\ years$

substitute in the formula

$V=15,300(1-0.1425)^{6}$

$V=15,300(0.8575)^{6}$

$V=\$6,082.70$

3 0

2 years ago

Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I

xxTIMURxx [149]

Answer:

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

$p_v =2*P(t_{(1699)}>1.971)=0.049$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=669$ represent the sample mean

$s=732$ represent the sample standard deviation

$n=1700$ sample size

$\mu_o =68$ represent the value that we want to test

$\alpha=0.$ 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 is different from 634, the system of hypothesis would be:

Null hypothesis: $\mu = 634$

Alternative hypothesis: $\mu \neq 634$

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971$

P-value

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

$df=n-1=1700-1=1699$

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

$p_v =2*P(t_{(1699)}>1.971)=0.049$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0

1 year ago

An object is moving at a speed of 850 miles per hour. Express this speed in meters per minute. Round your answer to the nearest

lorasvet [3.4K]

Answer:it would be 1385

Step-by-step explanation:because 850 miles per hour converted into meter per min would be 22799.04 meters per sec and then u would Find the number in the whole number place

4

and look one place to the right for the rounding digit on the right side of the decimal point

7

.Round up if this number is greater than or equal to

5

and round down if it is less than

5

1385

5 0

1 year ago

The volume inside of a sphere is V=4πr33 where r is the radius of the sphere. Your group has been asked to rearrange the formula

Semmy [17]

Answer:

Therefore,

$r=\sqrt[3]{\frac{3V}{4\pi }}$

is the required r

Step-by-step explanation:

Given:

Volume of inside of the sphere is given as

$V=\frac{4}{3} \pi r^{3}$

where r is the radius of the sphere

To Find:

r =?

Solution:

We have

$V=\frac{4}{3} \pi r^{3}$ ......Given

$3\times V=4\pi r^{3} \\\\\therefore r^{3}=\frac{3V}{4\pi } \\\\\therefore r=\sqrt[3]{\frac{3V}{4\pi }} \textrm{which is the expression for r}$

Therefore,

$r=\sqrt[3]{\frac{3V}{4\pi }}$

is the required r

5 0

1 year ago

For a set of data: x = (0,1,2,3,4,5,6) and y=(36, 28, 25, 24, 23, 21, 19), is it wise to use a linear regression to extrapolate

melisa1 [442]

Answer:

The problem with this solution is that a regression model is not recommended to extrapolate because we do not know if the linear relation that we calculated for a specific range of x values still holds outside this range.

Step-by-step explanation:

We have a linear regression model, with a range of the independent variable "x" that goes from 0 to 6.

The regression model finds a good fit (r=0.8582).

As it has a good fit, it is proposed to use this model to extrapolate and calculate the value of y for x=50.

It is not recommended to extrapolate a regression model unless we are really sure that the model is still valid within the range within we are extrapolating.

This means that if we have no proof that y has a linear relation in a range of x that includes x, the extrapolation has no validity and can lead to serious errors.

A linear regression model is only suitable for interpolation or extrapolating within the range we are sure that the relation between y and x is linear within a certain acceptable error.

4 0

1 year ago

Solve for the roots in the following equation. hint: factor both quadratic expressions. (x 4 5x 2 - 36)(2x 2 9x - 5) = 0