How much would $500 invest at 4% interest compounded annually be worth after 7 years?

Question Help

Alborosie

Answer:

42 l think l hope lm right

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Plz help this is due today AND NO LINKS OR GROSS PICTURES OR I REPORT

Amiraneli [1.4K]

0, scalene triangle. none of the sides are the same length

6 0

3 years ago

Erythromycin is a drug that has been proposed to possibly lower the risk of premature delivery. A related area of interest is it

katrin2010 [14]

Answer:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

Step-by-step explanation:

Data given and notation

n=195 represent the random sample taken

X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy

$\hat p=\frac{65}{195}=0.333$ estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy

$p_o=0.3$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:

Null hypothesis: $p\leq 0.3$

Alternative hypothesis: $p > 0.3$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.333 -0.3}{\sqrt{\frac{0.3(1-0.3)}{195}}}=1.01$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.01)=0.156$

So the p value obtained was a very low value and using the significance level asumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%

6 0

3 years ago

F(y) = y² + 10y + 21 <br>f'(2) = ? <br>

Ahat [919]

Answer:

14

Step-by-step explanation:

f'(2) = 2y + 10

f'(2) = 2(2) + 10

= 4 + 10

= 14

3 0

2 years ago

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Use the graph below for this question:

skelet666 [1.2K]

The average rate of change of a function f(x) on the interval [a,b] is given by:

f(b) - f(a)

-------------

b-a

In this specific case, a=2 and b = 3.

f(2) = -3 and f(3) = -1.

Thus, the average rate of change here of f(x) on the interval [2,3] is:

-1 - (- 3) 2

------------- = ------------- = 2 (answer)

3-2 1

5 0

3 years ago