Alicia borrowed $15,000 to buy a car. She borrowed the money at 8% for 6 years. What is the interest she will pay for the loan ?

PLEASE HELP ME WILL MARK BRAINLIEST! 10 POINTS!!!!!!!!!!

Darya [45]

Yes his equations are correct

3 0

2 years ago

In the figure, AngleRQS Is-congruent-to AngleQLK. 3 lines are shown. Lines S P and K N are parallel. Line R M intersects line S

stepan [7]

Answer:

x=108

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠RQS≅m∠QLK -----> by corresponding angles

m∠KLM+m∠QLK=180° -----> by supplementary angles (consecutive interior angles)

we have that

m∠RQS=x° ----> given problem

so

m∠QLX=x°

m∠KLM=(x-36)° ----> given problem

substitute

$(x-36)\°+x\°=180\°\\2x=180+36\\2x=216\\x=108$

3 0

2 years ago

Read 2 more answers

If ce = 7x+4, find the value of x

Alex787 [66]

Hey there,

To find the value of x we can take the two, CD or DE and set them equal to CE.

First step is set the lines equal to each other:

x + 3 + 8x - 9 = 7x + 4

Now we combine like terms:

9x - 6 = 7x + 4

Then we subtract 7x from both sides:

2x - 6 = 4

Add 6 to both sides:

2x = 10

Lastly we divide both sides by 2:

x = 5

So the value of x would have to be 5.

Hope I helped,

Amna

7 0

2 years ago

Read 2 more answers

Suppose that you bet $5 on each of a sequence of 50 independent fair games. Use the central limit theorem to approximate the pro

amm1812

Answer:

chances chances of happening = 0.0119

Step-by-step explanation:

given data

bet = $5

independent fair games = 50

solution

we will think game as the normal distribution

so here mean is will be

mean = $\frac{50}{2}$

mean = 25

and standard deviation will be

standard deviation = $\sqrt{50*0.5*0.5}$

standard deviation = 3.536

so

we have to lose 33 out of 50 time for lose more than $75

so as chance of doing things z score is

z score = $\frac{33-25}{3.536}$

z score = 2.26

so from z table

chances chances of this happening = 0.0119

8 0

2 years ago

We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys

marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation

$\bar X=19.2$ represent the battery life sample mean

$\sigma=2.5$ represent the population standard deviation

$n=25$ sample size

$\mu_o =18$ represent the value that we want to test

$\alpha$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:

Null hypothesis: $\mu = 18$

Alternative hypothesis: $\mu \neq 18$

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis: $\mu \leq 18$

Alternative hypothesis: $\mu > 18$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test for parts I and II, the p value would be:

$p_v =2*P(t_{(24)}>2.4)=0.0245$

And for part III since we have a one right tailed test the p value is:

$p_v =P(t_{(24)}>2.4)=0.0122$

5) Conclusion

I. This finding is significant for a two-tailed test at .05.

Since the $p_v$ . We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

6 0

2 years ago