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den301095 [7]
4 years ago
13

How can estimation help me to find the area of a rectangle or square?

Mathematics
1 answer:
faust18 [17]4 years ago
5 0

Answer:

You can identify whether your answer is correct or not and/or easily solve for an answer.

Step-by-step explanation:

To see if things make sense, estimating an answer can give you that clarification. More so, in high school geometry, you tend to get lots of decimals and radicals and all sorts of crazy numbers so estimation can really help. I hope that helps you. If that was not what you were looking for please feel free to ask.

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Simplify the following expression 2√2 x √12
MAVERICK [17]

Answer:

4  4√3

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
9y + 10 = -8<br><br> Please explain how you got your answer
Ronch [10]

You are trying to solve for y (get y by itself)

9y + 10 = -8       Subtract 10 on both sides

9y + 10 - 10 = -8 - 10

9y = -18         Divide 9 on both sides

y = -2

8 0
3 years ago
Find the Length of arc . Round to the nearest hundredth.
pshichka [43]
A) 14.66

Arc length is equal to radius multiplied by the central angle in radians not degrees

4 0
3 years ago
A simple random sample of 81 8th graders at a large suburban middle school indicated that 87% of them are involved with some typ
vodka [1.7K]

The 90% confidence interval that estimates the proportion of 8th graders that are involved in an after-school activity is (0.80853,0.93147).

The confidence interval for a population proportion for a given sample is given by the formula:

(p-Z\sqrt{\frac{p(1-p}{n} } , p+Z\sqrt{\frac{p(1-p}{n} }),

where p is the population proportion, Z is the Z-score value for the confidence interval, and n is the sample size.

In the question, we are given a random sample of 81 8th graders at a large suburban middle school. This implies that the sample size, n = 81. Also, we are told that they indicated 87% of them were involved with some type of after-school activity. This implies that the population proportion, p = 87% = 0.87.

We are asked to find the 90% confidence interval that estimates the proportion of them that are involved in an after-school activity.

Z-score (Z) corresponding to a 90% confidence interval is 1.645.

Thus, we can find the confidence interval as follows:

(0.87 - 1.645√{(0.87(1 - 0.87))/81},0.87 + 1.645√{(0.87(1 - 0.87))/81})

= (0.87 - 0.061469,0.87 + 0.061469)

= (0.80853,0.93147).

Thus, the 90% confidence interval that estimates the proportion of 8th graders that are involved in an after-school activity is (0.80853,0.93147).

Learn more about confidence intervals of population proportions at

brainly.com/question/13950323

#SPJ4

7 0
2 years ago
Your average weekly take home wage is $615. You take a one-week paid vacation and a second week unpaid vacation. You have calcul
pshichka [43]

Answer: $45

Step-by-step explanation:

As for the information provided,

The weekly rate of wages = $615

In a year there are 52 weeks, thus annual wages = $615  52 = $31,980

But since one week is an unpaid leaves, there will be no wages for such week.

Accordingly, annual wages shall be reduced by wages of a week = $31,980 - $615 = $31,365

Note: When the leaves are paid leaves wages are earned for such period.

Also the annual expected expenses = $31,320

Therefore, expected surplus at year end = $31,365 - $31,320 = $45

6 0
3 years ago
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