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zmey [24]
4 years ago
8

Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people ans

wered that they generally sell five cars; nineteen generally sell six cars; twelve generally sell seven cars; nine generally sell eight cars; eleven generally sell nine cars. Calculate the following. (Round your answer to two decimal places.)
sample mean = x

x = _____ cars
Mathematics
1 answer:
mr_godi [17]4 years ago
5 0

Answer:

x = 6.75 cars

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 65

14 people generally sell 5 cars

19 generally sell 6 cars

12 generally sell 7 cars

9 generally sell 8 cars

11 generally sell 9 cars

We have to calculate the sample mean.

Formula:

\text{Weighted mean} = \displaystyle\frac{\sum w_ix_i}{\sum w_i}\\\\\sum w_ix_i =  14(5) + 19(6) + 12(7) + 9(8) + 11(9) = 439\\\\\sum w_i = 14 + 19 + 12 + 9 + 11 = 65\\\\\text{Weighted mean} = \frac{439}{65} = 6.75

Thus, the weighted sample mean is 6.75 cars.

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Answer:

it is 19

Step-by-step explanation:

follow the order of operations

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3 years ago
Please help it is due tomorrow.
svlad2 [7]
R = radius of basketball
r = radius of tennis ball 
R = 3.6r cube both sides
the ratio of the radii is 1: 3.6
cube both sides to find the ratio of the volumes 
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3 years ago
Kristen filled his 52 ounce mug full of his favorite soda at the gas station it cost a total of $1.07 how much money is it for e
NikAS [45]

Answer:

$0.02

Step-by-step explanation:

$1.07 / 52 = $0.0205769230769

Simplified is $0.02

4 0
3 years ago
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How to solve this :') please help
blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

so

\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}

7 0
2 years ago
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