What is the area of this figure? :)
What is the area of this figure? :) - 1

Answers

Answer 1
Answer: there are 4 semicircles, each radius is 8cm, so we get the total areas of these semicircles: 4×0.5×8^(2)π
and the square area: 16²
so the total area of figure: 16²+4×0.5×8^(2)π=256+128π

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Please help me don’t know how to do it,
Which of the following ratios is not equivalent to 1.22?

Pls help due tomorrow just numbers 7 - 12

Answers

Answer:

Step-by-step explanation:

7). 5.75% = (5.75)/(100) = (575)/(10000) = (23)/(400)

10). 16(2)/(3) % = (50)/(3) % = (50)/(300) = (1)/(6)

12). 79(5)/(6) % = (479)/(6) % = (479)/(600)

Now is you turn. You can do it!

I need help with math ​

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How can I help you?

The sixth grade class is ordering special t-shirts. There are 88 sixth graders in all. The only sizes they can select are small, medium, or large. The class needs five times as many medium shirts as small shirts, and twice as many large shirts as small shirts. Let s represent the number of small shirts. Use any method (algebraically, tables, or tape diagrams) to show your work and solve for how many small, medium, and large shirts the class will need to order.

Answers

Answer:

166,000

Step-by-step explanation:

1x166,000=166,000

Hello I am needing help on this word problem please. I think it's 30 but I'm questioning myself

Answers

Solution:

Since 5 pounds of meat will feed about 26 people.

Thus;

\begin{gathered} (5)/(26)lb\text{ would feed 1 person} \n  \n =0.1923 \end{gathered}

If she is expecting 156 people, she should prepare;

\begin{gathered} (0.1923*156)lb\text{ of meat} \n  \n \approx30lb \end{gathered}

ANSWER: 30lb

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.912 + 20t + 65. What is theobject's maximum height?

Answers

Answer:

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

     a=-4.9   b=20  c=65

h(2) = -4.9(2)² + 20(2) + 65

     = -19.6 + 40 + 65

     = 85.4

Step-by-step explanation:

The weights, in pounds, of the cats in an animal adoption center are normally distributed. If a random sample of cats is taken and the confidence interval is (7.9,12.7), what is the margin of error

Answers

Answer: 2.4

Step-by-step explanation:

Given : The weights, in pounds, of the cats in an animal adoption center are normally distributed.

The confidence interval is (7.9,12.7)           (1)

Let \overline{x} be the sample mean of the weights, in pounds, of the cats in an animal adoption center.

We know that the confidence interval for population mean is given by :-

(\overline{x}-E,\overline{x}+E)      (2)

From (1) and (2)

\overline{x}-E=7.9-------(3)\n\n\overline{x}+E=12.7--------------(4)

Subtracting (3) from (4), we get

2E=4.8\n\n\Rightarrow\ E=2.4

Hence, the margin of error is E = 2.4