Answer: 13 sheets of paper
Step-by-step explanation:
We are given the dimensions of the box and the wrap paper:
Box:

Warp paper:

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




Warp paper:

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

This means Angel's dad needs to purchase 13 sheets of wrapping paper.
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
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.
8cm, since 6 full squares and 4 halves
hope it helps...!!!
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

Compute the sample sizes as follows:



Thus, the sample sizes are 6787.
(b)
Now it is provided that:

Compute the sample size as follows:

![n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%28z_%7B%5Calpha%2F2%7D%29%5E%7B2%7D%5Ctimes%20%5B%5Chat%20p_%7B1%7D%281-%5Chat%20p_%7B1%7D%29%2B%5Chat%20p_%7B2%7D%281-%5Chat%20p_%7B2%7D%29%5D%7D%7BMOE%5E%7B2%7D%7D)
![=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.33%5E%7B2%7D%5Ctimes%20%5B0.45%281-0.45%29%2B0.58%281-0.58%29%5D%7D%7B0.02%5E%7B2%7D%7D%5C%5C%5C%5C%3D6665.331975%5C%5C%5C%5C%5Capprox%206666)
Thus, the sample sizes are 6666.