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kirza4 [7]
1 year ago
15

Determine whether the relationship is an inverse variation or not. Explain.

Mathematics
1 answer:
kolbaska11 [484]1 year ago
8 0

Answer: The product xy is not constant, so the relationship is not an inverse variation.

Step-by-step explanation:

For it to be an inverse variation, the product of x and y must be constant.

  • For the first row, xy=2(424)=848.
  • For the second row, xy=3(280)=840.
  • For the third row, xy=4(210)=840.

This means <u>the product xy is not constant, so the relationship is not an inverse variation.</u>

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Jeri finds a pile of money with at least $\$200$. If she puts $\$50$ of the pile in her left pocket, gives away $\frac23$ of the
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Answer:

The possible values of the number of dollars in the original pile of money is ≥ $200 but < $350

Step-by-step explanation:

Here we have, pile of money ≥ $200

Amount in put the left pocket = $50

Fraction given away = 2/3 of rest of pile ≥ 2/3×150 ≥ $100

Amount put in right pocket = ≥ $150 - $100 ≥ $50

Total amount remaining with Jeri = $50 +≥ $50 ≥ $100

Also original pile - $200 < $100

Therefore where maximum amount given away to have more money = $200 we have

2/3× (original pile - 50) = $200

Maximum amount for original pile = $350

Therefore the possible values of the number of dollars in the original pile of money is ≥ $200 but < $350.

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Find the volume of the square pyramid. Round your answer to the nearest hundredth.
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B

Step-by-step explanation:

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Preliminary data analyses indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the
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<h2>Answer with explanation:</h2>

By considering the given information, we have

Null hypothesis : H_0: \mu=35.0

Alternative hypothesis : H_1: \mu\neq35.0

Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.

\overline{x}=41.0    ;    \sigma= 3.7

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

i.e. t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25

Degree of freedom : n-1 = 20-1=19

Significance level = 0.01

For two tailed,  Significance level =\dfrac{0.01}{2}=0.005

By using the t-distribution table, the critical value of t =t_{19, 0.005}=2.861

Since , the observed t-value (7.25) is greater than the critical value (2.861) .

So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.

We conclude that there is some significance difference between the mean score for sober women and 35.0.

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A

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