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tatuchka [14]
2 years ago
5

So I really need help. you have to set fractions equal to each other and try and find out how many prefer sweaters and whatnot.

Also, the answer isn't supposed to be a fraction, it's supposed to be a whole number, you find the answer by using fractions with the information provided in the word problem above ^
I hope that makes sense. lol, thank you :)​

Mathematics
2 answers:
soldi70 [24.7K]2 years ago
6 0

Answer:

84

Step-by-step explanation:

52%of math problem are pointless

52%of math problem = pointless

52% of 175

(52/100)×175

91 =pointless

math problem - pointless =not pointless

175 - 91 = 84

beks73 [17]2 years ago
5 0
52/100 X 175



Solve it
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G(x)= 3x − 3; Find g(−6)
Irina18 [472]
<span>g(x)= 3x − 3
</span><span>g(−6)

g(-6) = 3(-6) - 3
g(-6) = -18 - 3
g(-6) = -21

The answer is -21.</span>
4 0
3 years ago
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If an area can be washed at a rate of 3,600 cm2/minute, how many square inches can be washed per hour? Enter your answer with th
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Answer:

276923 square inches

Step-by-step explanation:

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3 years ago
Just need help with #2 <br> Please help!<br> Thank you <br> (15 points)
IrinaK [193]
So you write into the table that x = 1 then y will equal 1*1.5 which equal to 1.5. So under x = 1 in y table you write y = 1.5, and so on

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What is the value of the third quartile of the data set represented by this box plot?
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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
2 years ago
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