HELP OFFERING HUGE POINTS!!!!!!!!

noname [10]

1790988582563.6448is your answer. All you do is multiply 14 times 1.2, which equals 16.8. Then you do 16.8 times itself. 16.8 x 16.8 = 282.24. Then 282.24 times itself, and so on so forth.

8 0

4 years ago

Ms. Whodunit needs $15 000 to go on her dream vacation in four years. How much

lisov135 [29]

For each situation, we have that:

11) Using compound interest, it is found that she needs to invest $9,781.11 now.

12) Using the future value formula, it is found that you will have $728,753 after 48 years.

<h3>What is compound interest?</h3>

The amount of money earned, in compound interest, after t years, is given by:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

In which:

A(t) is the amount of money after t years.

P is the principal(the initial sum of money).

r is the interest rate(as a decimal value).

n is the number of times that interest is compounded per year.

For this problem, the parameters are given as follows:

A(t) = 15000, t = 4, r = 0.055, n = 2.

Hence we solve for P to find the amount that needs to be invested.

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$15000 = P\left(1 + \frac{0.055}{2}\right)^{2 \times 8}$

$(1.0275)^{16}P = 15000$

$P = \frac{15000}{(1.0275)^{16}}$

P = $9,718.11.

She needs to invest $9,781.11 now.

<h3>What is the future value formula?</h3>

It is given by:

$V(n) = P\left[\frac{(1 + r)^{n-1}}{r}\right]$

In which:

P is the payment.

n is the number of payments.

r is the interest rate.

For item 12, the parameters are given as follows:

P = 150, r = 0.07/12 = 0.005833, n = 48 x 12 = 576.

Hence the amount will be given by:

$V(n) = P\left[\frac{(1 + r)^{n-1}}{r}\right]$

$V(n) = 150\left[\frac{(1.005833)^{575}}{0.005833}\right]$

V(n) = $728,753.

You will have $728,753 after 48 years.

More can be learned about compound interest at brainly.com/question/25781328

#SPJ1

6 0

2 years ago

Show that (a+b)^2=4p^2

Alja [10]

LHS :-

( a + b)²
(P + q + P - q)²
(2p)²
2²p²
4p² = RHS

8 0

3 years ago

Write the fractional notation in lowest terms. <br> 110 minutes to 4 hours.

Angelina_Jolie [31]

It will be 27 mins to 2 hours I think

5 0

3 years ago

An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty i

grin007 [14]

Answer:

Step-by-step explanation:

The question is incomplete. The missing information is the group of answer choices. The group of answer choices are

a) none of the above

b) Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).

c) Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).

d) Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).

e) Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).

Solution:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ ≥ 10

For the alternative hypothesis,

µ < 10

This is a left tailed test.

Since the number of samples is small and the population standard deviation is not given, the distribution is a student's t.

Since n = 9,

Degrees of freedom, df = n - 1 = 9 - 1 = 8

t = (x - µ)/(s/√n)

Where

x = sample mean = 13.5

µ = population mean = 10

s = samples standard deviation = 3.2

t = (13.5 - 10)/(3.2/√9) = 3.28

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a left tailed test. Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.01 = 0.99

The negative critical value is - 2.896

Since - 3.28 is lesser than - 2.896, then we would reject the null hypothesis.

By using probability value,

We would determine the p value using the t test calculator. It becomes

p = 0.0056

Level of significance = 1%

Since alpha, 0.01 > than the p value, 0.0056, then we would reject the null hypothesis. Therefore, At a 1% level of significance, the sample data showed significant evidence that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly less than 10 years

The correct option is

a) none of the above

8 0

3 years ago