PLEASE HELP! The AHS football team did a weigh-in at the start of training camp. The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg. What percentage of players are under 86kg?

PLEASE HELP! The AHS Football Team Did A Weigh-in At The Start Of Training Camp. The Weights Of The Players

Answers

Answer 1
Answer:  2.5%

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Explanation:

Notice how 98-2*6 = 98-12 = 86. So we're 2 standard deviations below the mean.

Start at the center, which is where the highest point is located. The value 98 is at the center. Then move 2 spots to the left, which has us move 2 standard deviations lower than the mean, and we arrive at 86. Refer to the diagram below.

From here, we add up the percentages in that diagram that are below the 86 mark. Those percentages are the 0.15% and 2.35%

So we get (0.15%) + (2.35%) = 2.5% which is the answer

Side note: This diagram is using the Empirical Rule. Some textbooks call it the "68-95-99.7 rule", but that's more clunky of a name in my opinion. The three values 68, 95 and 99.7 refer to the approximate percentage values within 1, 2 and 3 standard deviations of the mean.

PLEASE HELP! The AHS Football Team Did A Weigh-in At The Start Of Training Camp. The Weights Of The Players

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Jorge has $150 in his savings account. He earns $33 a week mowing lawns. If Jorge saves all of his earnings, after how many weeks will he have$437 saved?

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Answer:

9 weeks

Step-by-step explanation:

used a calculator really easy

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Jason inherited a piece of land from his great-uncle. Owners in the area claim that there is a 45% chance that the land has oil. Jason decides to test the land for oil. He buys a kit that claims to have an 80% accuracy rate of indicating oil in the soil. What is the probability that the land has oil and the test predicts it

Answers

Answer:

0.36 = 36% probability that the land has oil and the test predicts it

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

45% chance that the land has oil.

This means that [tex]P(A) = 0.45[/tex]

He buys a kit that claims to have an 80% accuracy rate of indicating oil in the soil.

This means that [tex]P(B|A) = 0.8[/tex]

What is the probability that the land has oil and the test predicts it?

This is [tex]P(A \cap B)[/tex]. So

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(B \cap A) = P(B|A)*P(A) = 0.8*0.45 = 0.36[/tex]

0.36 = 36% probability that the land has oil and the test predicts it

Answer:

The probability that the land has oil and the test predicts it is 36%

Step-by-step explanation:

So option  C.  0.36 is correct for plato users

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the answer is 17 because that’s how many players ran over 1 score

Answer:

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Step-by-step explanation:

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8 people scored 3

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True
False

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Answer:

False

Step-by-step explanation:

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Answers

Answer:

Option E, two-proportion z test  should be used to determine whether these data provide sufficient evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same as the proportion of shoppers in the large downtown shopping area who had been to a movie in the past month

Step-by-step explanation:

The complete question is

In a random sample of 60 shoppers chosen from the shoppers at a large suburban mall, 36 indicated that they had been to a movie in the past

month. In an independent random sample of 50 shoppers chosen from the shoppers in a large downtown shopping area, 31 indicated that

they had been to a movie in the past month. What significance test should be used to determine whether these data provide sufficient

evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same

as the proportion of shoppers in a large downtown shopping area who had been to a movie in the past month?

A one-proportion z interval B two-proportion z interval

B two-proportion z interval

C two-sample t test D one-proportion z test

D one-proportion z test

E two-proportion z test

Solution

Two proportion z test is used to compare two proportions. In this test the null hypothesis is that the two proportions are equal and the alternate hypothesis is that the proportions are not the same. The random sample of populations serve as two proportions.

Hence, option E is the best choice of answer

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

Hello There!

The first thing we need to do is find the formula for area of a trapezoid

The formula is

[tex]A=\frac{a+b}{2} h[/tex]

where a= base 1

b = base 2

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we are given all of the dimensions so all we have to do is plug them in

sp

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32 units ²

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32

Step-by-step explanation:

Find formula

A=a+b/2 h

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Answers

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Answer:

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Step-by-step explanation:

10x + 4 - 6x + 1

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Answer:

Step-by-step explanation:

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Answer:

b = 11/9 = 1 2/9

Step-by-step explanation:

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