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umka2103 [35]
2 years ago
11

Earned income and capital gains (or "portfolio income") are acquired in different ways. Which statement describes how they are d

ifferent?
Mathematics
1 answer:
Dvinal [7]2 years ago
5 0

Answer:

n

Step-by-step explanation:

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bagirrra123 [75]

Answer:

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Step-by-step explanation:

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2 years ago
A rancher has 300 feet of fencing to enclose a pasture bordered on one side by a river. the river side of the pasture needs no f
Mashcka [7]
I think your best answer is a pasture that is 75 feet by 150 feet, which will give you 11,250 sq ft of pasture
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3 years ago
Find the median. 90, 88, 100, 82, 80.
gavmur [86]

Answer:

100 is the median for te above observations

<h2><em><u>HOPE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>HELPED</u></em><em><u> </u></em></h2>

3 0
3 years ago
Read 2 more answers
The area of a trapezium is 360 cm² and its height is 18 cm. If one of the parallel sides is longer than the other by 6 cm, then
Leokris [45]

Answer:

The two parallel sides a and b are 17 cm and 23 cm respectively

Step-by-step explanation:

Area of a trapezium = {(a+b)/2} h

Where

h = height = 18 cm

a = parallel side 1 = x

b = parallel side 2 = x + 6

Area of a trapezium = {(a+b)/2} h

360 = {(x + x + 6) / 2} 18

360 = { (2x + 6) / 2} 18

360 = (2x + 6) 9

360 = 18x + 54

Subtract 54 from both sides

360 - 54 = 18x

306 = 18x

Divide both sides by 18

x = 306 / 18

= 17

x = 17 cm

a = parallel side 1 = x

x = 17 cm

b = parallel side 2 = x + 6

x + 6

= 17 + 6

= 23 cm

8 0
2 years ago
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions 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.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

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

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

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

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

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

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

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

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

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

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

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

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
2 years ago
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