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AleksAgata [21]
2 years ago
7

John spent $75 on a shopping trip for new clothes last week. He had expected to spend $100 on clothes. Complete the table.John's

sister went shopping with him. She expected to spend $100 on clothes, but spent only $10. John wanted to help her calculate the percent error, but he thinks he may have made a mistake. Are his calculations below correct? Why or why not?
Johns Calculations: 100-10 10
--------- = ----- is approximately 11%
90 90
Mathematics
2 answers:
aleksklad [387]2 years ago
5 0
John’s calculations are not correct. He compared the exact value to the absolute error. The absolute error should be in the numerator and the exact value in the denominator.
Gala2k [10]2 years ago
5 0

Answer:

Amount Possessed by John's sister = $ 100

Amount spent on buying clothes = $ 10

So, % of money spent on buying clothes by john's sister

         = \frac{\text{ Amount spent by john's sister}}{\text{Amount possessed by john's sister }}\times 100=\frac{10}{100}\times{100} = 10%

Johns Calculations: 100-10 10

--------- = ----- is approximately 11%

90 90 is incorrect.


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Let the number of sports drinks April bought be represented by = s

Let the pizza slices April bought be represented by = p

As she spent a total of $21.25 so equation becomes:

s+p=21.25

p+3+p=21.25

2p=18.25

p=9.125

As she bought 3 more sports drinks than pizza slices so, s = p+3

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Find f(x) and g(x) so that the function can be described as y = f(g(x)). y = nine divided by square root of quantity five x plus
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y=\frac{9}{\sqrt{5x+5}}
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f(x)=\frac{9}{x} and
g(x)=\sqrt{5x+5}

we could also have
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or
f(x)=\frac{9}{\sqrt{x+5}} and
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For the first year of his car insurance policy, Tom pays $77 per month. Tom's payments are reduced to $65 per month in the secon
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Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes
dimaraw [331]

Answer:

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Step-by-step explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let \mu = <u><em>population mean length of the songs</em></u>

So, Null Hypothesis, H_0 : \mu \leq 210 seconds      {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis, H_A : \mu > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;

                              T.S.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean length of songs = 252.5 seconds

            \sigma = population standard deviation = 54.5 seconds

            n = sample of current hit songs = 40

So, <u><em>the test statistics</em></u> =  \frac{252.5-210}{\frac{54.5}{\sqrt{40} } }

                                   =  4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

4 0
2 years ago
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