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2 years ago

I really don’t know what to do someone help ASAP

Mathematics

2 answers:

Archy [21]2 years ago

5 0

Answer:

0.001

Step-by-step explanation:

-BARSIC- [3]2 years ago

3 0

Answer:

5.5

Step-by-step explanation:

The volume of a sphere is represented by the formula 4/3πr^3. So, if we set 4/3πr^3 equal to 2048/3, we can start solving. Start by multiplying both sides by 3/4, the reciprocal of 4/3 to get rid of the fraction. This gives us πr^3 = 512. Divide 512 by π to get 163.1 (when rounded to the nearest tenth). This leaves us with r^3 = 163.1; find the cubic root of 163.1, which is 5.5, to get r

(a more exact answer is 5.4631958645870141851784948846314‬)

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A farmer wants to fence in a rectangular field that encloses 3600 square feet. One side of the field is along a river and does n

andriy [413]

Answer:

C(x) = $\$3.50(\frac{x^2+7200}{x})$

Step-by-step explanation:

Data provided in the question:

Area of the field = 3600 square feet

Fencing charges = $3.50 per foot

Let the side along the river be 'x' feet and the other side of the 'B'

now,

Area of rectangle = Bx = 3600 square feet

B = $\frac{3600}{x}$ feet

and total length to be fenced = x + 2B

therefore,

Total cost of fencing = Fencing charges × total length to be fenced

Total cost of fencing = $3.50 × ( x + 2B )

substituting the value of B from (1)

Total cost of fencing, C(x) = $\$3.50(x+2\times\frac{3600}{x})$

C(x) = $\$3.50(\frac{x^2+7200}{x})$

4 0

3 years ago

"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen

Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± $Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ ]

Where

p' is the sample proportion/point estimator of the population proportion

$Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$ is the margin of error (d) of the confidence interval.

$Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648$

$d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }$

$d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }$

$(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}$

$n*(d*Z_{1-\alpha /2})^2= p'(1-p')$

$n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}$

$n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}$

n= 147.28 ≅ 148 boxes.

I hope it helps!

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Step-by-step explanation:

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Answer:

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Answer:

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