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frez [133]
4 years ago
10

An architect created plans for a house using a scale factor of 1:16 . In the plans, the floor of the house has an area of 7 squa

re feet. What is area of the floor in the actual house? Enter your answer in the box.
A rectangle has a length of 6 inches and a width of 3 inches.

What is the effect on the perimeter when the dimensions are multiplied by 8?

The perimeter is increased by a factor of 8.
The perimeter is increased by a factor of 24.
The perimeter is increased by a factor of 64.
The perimeter is increased by a factor of 256.

This figure is made up of a triangle and a semicircle.

What is the area of this figure?

Use 3.14 for pi. Round only your final answer to the nearest tenth.

Enter your answer, as a decimal, in the box.

Mathematics
1 answer:
charle [14.2K]4 years ago
7 0
Q1.the given scale factor is 1:16
the area is 7 square feet
when calculating the area the scale factor is squared as well. 
since the scale factor is 16, to find out area the scale should be squared that is,
16*16 = 256

then the new area is 7 square feet * 256 
actual area = 256 * 7 = 1792 square feet 
Q2. the rectangle with length 6 inches and width 3 inches. 
the perimeter of the rectangle before changing dimensions are;
=(6*2) + (3*2) 
= 12 + 6 = 18 inches
after multiplying the dimensions by 8,
new length = 6*8 = 48 inches
new width = 3*8 = 24 inches 
new perimeter = (48*2) + (24*2)
                        = 96 + 48 
                        = 144 inches
the perimeter ratio from new to old = 144:18
the perimeter has increased by (144/18) times
= 144/18 = 8
answer is perimeter increased by a factor of 8

q3) 
area of the triangle 
= 1/2 * height * base
= 1/2 * (5-2) *(4-(-3))
= 1/2 * 3 * 7
= 10.5 units²
area of semicircle 
=pi*r2
r = 2-(-1.5)
 = 3.5
area = 3.14 * (3.5)²
        = 38.465 units²
total area = area of triangle + area of semicircle 
               = 10.5 + 38.465
              = 48.965
round it off to the closest tenth
              area = 49 units²
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Answer:

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 1000, \sigma = 350, n = 50, s = \frac{350}{\sqrt{50}} = 49.5

What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

This is the pvalue of Z when X = 1000+50 = 1050 subtracted by the pvalue of Z when X = 1000-50 = 950. So

X = 1050

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

By the Central Limit Theorem

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

Z = \frac{1050-1000}{49.5}

Z = 1.01

Z = 1.01 has a pvalue of 0.8438

X = 950

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

Z = \frac{950-1000}{49.5}

Z = -1.01

Z = -1.01 has a pvalue of 0.1562

0.8438 - 0.1562 = 0.6876

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

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