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3 years ago

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I need help with this question

1 answer:

Ghella [55]3 years ago

3 0

${6}^{2} - {5}^{2} \\ 36 - 25 \\ 11$
the answer is 11

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A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric

allochka39001 [22]

I'm fairly certain that using that wire you can have the same area no matter what dimensions there are, but if you do the square root, or $\sqrt{}$ of 19,001,900 then you get 4359.11688. So your dimensions are 4359.11688 x 4359.11688. To get area, do length x width to get <span> $1.9*1 0^{7}$ .

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4 0

3 years ago

Use the t-distribution and the sample results to complete the test of the hypotheses. Use a significance level. Assume the resul

Ksenya-84 [330]

Answer:

Test statistic = 1.3471

P-value = 0.1993

Accept the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4

Sample mean, $\bar{x}$ = 4.8

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation, s = 2.3

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4\\H_A: \mu \neq 4$

We use two-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{4.8 - 4}{\frac{2.3}{\sqrt{15}} } = 1.3471$

Now, we calculate the p-value.

P-value = 0.1993

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.

3 0

3 years ago

A car rental costs $50 for the first day. Additional days cost $35 per day, unless the car is rented for 7 days or more, in whic

son4ous [18]

Answer:

$\$50+\$31.5x$

Step-by-step explanation:

Let

x------> the number of days

y----> the cost of renting a car

we know that

For $x$

$y=\$50+\$35x$

For $x\geq 7\ days$

The rate is equal to

$0.90*\$35=\$31.5$

so

$y=\$50+\$31.5x$

In this problem. the car has been rented for more than a week

therefore

$x> 7\ days$

The cost of renting a car is equal to

$y=\$50+\$31.5x$

6 0

3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in, and a standard deviation given by

kirza4 [7]

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : $\mu= 62.5\text{ in}$

Standard deviation : $\sigma = \text{2.5 in}$

a) The formula for z -score :

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Sample size = 1

For x= 63 in. ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2$

The p-value = $P(z$

$0.5792597\approx0.5793$

Thus, the probability is approximately = 0.5793

b) Sample size = 35

For x= 63 ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18$

The p-value = $P(z$

$= 0.8809999\approx0.8810$

Thus , the probability is approximately=0.8810.

6 0

3 years ago

Suppose your friend's parents invest $10,000 in an account paying 4% compounded annually. What will the balance be after 9 years

LiRa [457]

Answer:

$14233.12

Step-by-step explanation:

10,000(1+.04/1)^1(9)

5 0

3 years ago