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2 years ago

a circle with radious of 1cm sits inside a 11cm times 12 cm rectangle what is the area of the shaded region

Mathematics

2 answers:

Nitella [24]2 years ago

7 0

Answer: 128.86cm²

Step-by-step explanation:

The circle is inscribed in the rectangle. To find the shaded portion, subtract he area of the circle from the are of the rectangle.

Area of the rectangle = 11 x 12

= 132cm²

Area of the circle with radius of 1cm = πr²

= 3.142 x 1²

= 3.142cm²

Therefore , area of the shaded region = 132cm² - 3.142cm²

= 128.86cm²

Yuki888 [10]2 years ago

7 0

Answer:128.86

Step-by-step explanation: I got it correct on khan

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Question 2 (6 points)

alexdok [17]

Answer:

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Step-by-step explanation:

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2 years ago

A catering service offers 7 appetizers, 12 main courses, and 6 desserts. A banquet chairperson is to select 5 appetizers, 6 main

RideAnS [48]

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3 years ago

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aleksklad [387]

We know that we have to make a down payment of $1500 and finance the rest of $20000 at a 1.9% interest rate, making equal monthly payments for 5 years. Our first step to solve this problem would be to convert 5 years into months.

1 year = 12 months

12 * 5 = 60 months

Therefore, in 5 years there are 60 months.

Now lets solve this problem step by step.

Subtract the down payment from $20,000

$20000-$1500=$18500

Multiply the remaining number by the interest rate.

$18500 *1.9 = $35150

Divide 35150 by number of months in 5 years (60)

$35150 / 60 = $585

Therefore, you have to pay $585 per month.

6 0

2 years ago

Add (13a + 5b) + (a + 5b - 3)

Anni [7]

(13a + 5b) + (a + 5b - 3)

= 13a + 5b + a + 5b - 3

= 14a + 10b - 3

5 0

3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0

3 years ago