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2 years ago

280 m

Mathematics

1 answer:

Marianna [84]2 years ago

6 0

Answer: C. 1,183m

Step-by-step explanation:

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angel's dad bought a television for her completing all of her digital learning day lesson. he want to warp the gift. the box mea

atroni [7]

Answer: 13 sheets of paper

Step-by-step explanation:

We are given the dimensions of the box and the wrap paper:

Box:

$(56 in)(8 in)(36 in)$

Warp paper:

$(26 in)(17 in)$

Now we need to find the surface area of the box and the area of the wrap paper:

Box:

$surface_{1}=(56 in)(8 in)(2)=896 in^{2}$

$surface_{2}=(8 in)(36 in)(2)=576 in^{2}$

$surface_{3}=(56 in)(36 in)(2)=4032 in^{2}$

$surface-area=896 in^{2}+576 in^{2}+4032 in^{2}=5504 in^{2}$

Warp paper:

$area=(26 in)(17 in)=442 in^{2}$

Dividing the area of the box by the area of the paper:

$\frac{surface-area}{area}=\frac{5504 in^{2}}{442 in^{2}}=12.45 \approx 13$

This means Angel's dad needs to purchase 13 sheets of wrapping paper.

4 0

3 years ago

⅖ = ½ x<br> Explain your reasoning

fomenos

x = 4/5

2/5=1/2x

Step 1: Flip the equation.

1/2x=2/5

Step 2: Multiply both sides by 2.

2*(1/2x)=2*(2/5)x=4/5

4 0

2 years ago

A toy rocket is launched straight up from the top of a building. The function that models the height as a function of time is h(

kicyunya [14]

Answer: 50 ft

Step-by-step explanation:

The correct function is h(t)=-16t^2+200t+50.

Hi, to answer this question we simply have to replace t =0 in the function given:

h(t)=-16t^2+200t+50.

h (0) = -16t^2+200t+50.

h(0) = 50

When the time is 0, the rocket is at the top of the building.

The rocket was launched from a 50 ft height.

Feel free to ask for more if needed or if you did not understand something.

4 0

3 years ago

What is the area of the shape below

Tems11 [23]

8cm, since 6 full squares and 4 halves

hope it helps...!!!

7 0

2 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}$

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

2 years ago