Help pls im dont know it

A savings account accrues interest at a rate of 3.0% yearly. If someone opens an account with $2,500, how much money would the a

Sauron [17]

$3\%=\frac{3}{100}=0,03\\\\
First\ year:2500+2500*0,03=2500+75=2575\\\\Second\ year:\ 2575+2575*0,03=2575+77,25=2652,25\\\\Third\ year:\ \ 2652,25+2652,25*0,03=2652,25+79,5675=\\\\2731,8175\\\\ Fourth\ year: 2731,8175+2731,8175*0,03=2731,8175+81,954525\\\\=2813,772025\\\\Fifth\ year:\ 2813,772025+2813,772025*0,03=\\\\2813,772025+84,41316075=2898,18518575\$$

5 0

4 years ago

each class AT BRIARWOOD ELEMentary collected at least 54 cans during the food drive. If there were 29 classes in the school what

sertanlavr [38]

You multiply 54 by 29 which will leave you with 1566

6 0

3 years ago

Read 2 more answers

Joe Popoff, a collection agent, collected 90% of a debt of $5,600.00 that had been overdue 90 days. This collection rate was 5%

sergey [27]

<span>You are given the debt collected by Joe Popoff with a 90% of a debt of $5,600.00 that had been overdue 90 days. This collection rate was 5% more than the average collection rate for that agent. The agent charged 25% commission. You are asked to find the net proceeds.

Amount collected
= $5,600 * 0.90
= $5,040

Commision
= $5,040 * 0.25
= $1,260

Net Proceeds
= $5,040 - $1,260
= $3,780 </span>

7 0

3 years ago

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales

andrezito [222]

Answer:

$t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657$

$p_v =P(t_{(39)}>6.657)=3.17x10^{-8}$

If we compare the p value and the significance level given $\alpha=0.01$ 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 higher than 42 at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=44$ represent the sample mean

$s=1.9$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =42$ represent the value that we want to test

$\alpha=0.01$ 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 higher than 42, the system of hypothesis would be:

Null hypothesis: $\mu \leq 42$

Alternative hypothesis: $\mu > 42$

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:

$t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657$

P-value

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

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>6.657)=3.17x10^{-8}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ 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 higher than 42 at 1% of signficance.

3 0

3 years ago

a chinchilla weighs 2 pounds its weight is 1/8 the weight of a koala write an expression and solve it to show how much the koala

Wittaler [7]

So you can say that the weight of a koala here can be x.

So x * 1/8 = 2

2 being the chinchilla's weight and the 1/8 coming from the fact that the chinchilla weighs 1/8 of the weight of the koala.

So you can just divide both sides by 1/8 first.

x * 1/8 = 2

/ 1/8 / 1/8

Or,

x *1/8 * 8/1 = 2 * 8/1

x = 16

So the koala weighs 16 pounds.

6 0

3 years ago