Write the percent as a decimal.<br> 3.7%

You put $200 in a savings account. The account earns 2% simple interest

Black_prince [1.1K]

Answer:

interest total after 5 years is $20.82

Step-by-step explanation:

6 0

3 years ago

QUESTION 1 - 1 POINT

guajiro [1.7K]

Answer + Step-by-step explanation:

the inequality by finding the boundary line, then shading the appropriate area.

y > − 5/2x − 5

(anything in shaded region)

8 0

2 years ago

Simplify. (-2y^3)^3(-3x^2yz^2)^3

Alik [6]

=-8y^9*-27x^6y^3z^6

=216x^6y^12z^6

6 0

3 years ago

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

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

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

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

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

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

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago

Five people each write a positive integer on a piece of paper. If the average of these five integers is 8, what is the greatest

erik [133]

Answer:

Option D. $40$