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2 years ago

8

You receive $1,000 to put in the bank. You place it in an account that pays 4% annual interest compounded continuously. How much

will you have in 15 years? Round the answer to the nearest dollar.

1 answer:

Mariana [72]2 years ago

7 0

(1.04^15)(1000)= $1,800.94

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A child’s wading pool is 3.8 feet wide, 5.5 feet long, and 1.2 feet deep. How many gallons of water will the pool hold? ( Use 1

vichka [17]

First, we find the volume of the pool.

3.8*5.5*1.2= 25.08.

Then we convert the feet into gallons.
25.08*7.5 = 188.1

The pool will hold 188.1 gallons of water.

3 0

2 years ago

Suppose that the researchers wanted to estimate the mean reaction time to within 6 msec with 95% confidence. Using the sample st

Lisa [10]

Answer:

a) The 95% confidence interval would be given by (509.592;550.308)

b) n=523 rounded up to the nearest integer

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=530$ represent the sample mean for the sample

$\mu$ population mean

s=70 represent the sample standard deviation

n=48 represent the sample size (variable of interest)

Confidence =95% or 0.995

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are df=n-1=48-1=47

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,47)".And we see that $z_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$530-2.01\frac{70}{\sqrt{48}}=509.692$

$530+2.01\frac{70}{\sqrt{48}}=550.308$

So on this case the 95% confidence interval would be given by (509.592;550.308)

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Assuming that $\hat \sigma =s$

And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval is provided, $z_{\alpha/2}=1.96$ , replacing into formula (b) we got:

$n=(\frac{1.96(70)}{6})^2 =522.88 \approx 523$

So the answer for this case would be n=523 rounded up to the nearest integer

3 0

2 years ago

Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ

solong [7]

Answer:

The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

Step-by-step explanation:

We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

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2 years ago

Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = xe−x(1 + y) x ≥ 0 an

Yuri [45]

Answer:

For the given explanation we see that two life times are not independent

Step-by-step explanation:

probability for X (for x≥ 0)

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dy$

$-e^-^x\int\limits^\infty_0 {de^-^x^y} \,$

e⁻ˣ

Probability for X exceed 3

= $\int\limits^\infty_3 {f(x)dx$

= $\int\limits^\infty_3 {e^-^3 dx$

= $e^-^3$

probabilty for y≥ 0 is

$\int\limits^\infty_0 {xe^-^x^(^1^+^y^)} \, dx \\$

$\int\limits^\infty_0{-x/1+yd(e^-^x^(^1^+^y^))} \, =1/(1+y)² \\$

3 0

2 years ago

Alisa says it is easier to compare the numbers in set A than the numbers in set B

kvv77 [185]

I am not sure is this is right, but you could say that set A has a number in the millions, which makes in easier to set apart from the other one.

6 0

2 years ago

Read 2 more answers