Soit les expressions A et B connues par :

Danny wants to buy a truck in 4 years. He is going to put away $2,500.00 into his savings account that will pay him 6.75% intere

aniked [119]

Answer:

b $3,272.43

Step-by-step explanation:

A = p(1+r/n)^nt

Where

A= future value

P= principal = $2500

r= interest rate = 6.75% = 0.0675

n = number of periods = 12

t = time = 4 years

A = p(1+r/n)^nt

= 2500(1+0.0675/12)^12*4

= 2500(1+0.005625)^48

= 2500(1.005625)^48

= 2500(1.3089737859257)

= 3272.4344648144

Approximately

A= $3272.43

He will have $3272.43 to give as down payment in 4 years

4 0

3 years ago

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below. See Attached Excel for Data. Assume t

motikmotik

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

The amounts (in ounces) of randomly selected eight 16-ounce beverage cans are given below.

16.5, 15.2, 15.4, 15.1, 15.3, 15.4, 16, 15.1

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans and fill in the blanks appropriately.

A 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is ( , ) ounces. (round to 3 decimal places)

Answer:

$99\% \: \text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

Step-by-step explanation:

Let us find out the mean amount of the 16-ounce beverage cans from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

Let us find out the standard deviation of the 16-ounce beverage cans from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.4957$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where $\bar{x}$ is the sample mean and Margin of error is given by

$$ MoE = t_{\alpha/2} \cdot (\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample size, s is the sample standard deviation and is the t-score corresponding to a 99% confidence level.

The t-score corresponding to a 99% confidence level is

Significance level = α = 1 - 0.99 = 0.01/2 = 0.005

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.005 and DoF = 7

t-score = 3.4994

$MoE = t_{\alpha/2}\cdot (\frac{s}{\sqrt{n} } ) \\\\MoE = 3.4994 \cdot \frac{0.4957}{\sqrt{8} } \\\\MoE = 3.4994\cdot 0.1753\\\\MoE = 0.6134\\\\$

So the required 99% confidence interval is

$\text {confidence interval} = \bar{x} \pm MoE\\\\\text {confidence interval} = 15.5 \pm 0.6134\\\\\text {confidence interval} = 15.5 - 0.6134, \: 15.5 + 0.6134\\\\\text {confidence interval} = (14.886, \: 16.113)\\\\$

Therefore, the 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (14.886, 16.113) ounces.

8 0

4 years ago

1=5<br>2=12<br>3=39<br>4=148<br>5=?

tankabanditka [31]

The answer might be 305?

4 0

3 years ago

What is the slope of the line?

Scorpion4ik [409]

Answer:

Slope is technically 1, because it implies that it's going up one and over to the right one. Also in slope-intercept form it would look like this

y = x + 1.

Step-by-step explanation:

Hope this makes sense :)

4 0

3 years ago

Jill has a poster hanging in her room whose width is 212 in. shorter than its length. The perimeter of the poster is 103 in. How

zhuklara [117]

The poster is a rectangle.
P = 2 L + 2 W
where L is the length and W is the width of a rectangle.
L = W + 2 1/2 = W + 2.5
103 = 2 ( W + 2.5 ) + 2 W
103 = 2 W + 5 + 2 W
103 = 4 W + 5
4 W = 103 - 5
4 W = 98
W = 98 : 4
W = 24.5 in
L = 24.5 + 2.5
L = 27 in
Answer:
The poster is 27 inches long and 24.5 inches wide.

4 0

3 years ago