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lisov135 [29]
3 years ago
8

I need help, i cant do this easly i suck at math so was wandering if yall can give me an answer for these or tell me how to do i

t.
I NEED HELP ASAP

Mathematics
2 answers:
AVprozaik [17]3 years ago
8 0
This is what I got from Photomath
Scorpion4ik [409]3 years ago
4 0
On all of them sorry but if i dont know wich one to do then....
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there would be 40,960 tree frogs i think

Step-by-step explanation:

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The expression
swat32

Answer:

The expression

1

÷

1/4

is given. Give a real-life application to explain this expression and then simplify it.

Given a life expression to solve this, when one divide 100 pieces of pencil by 25 which is 1/4 of the entire pencil.

1/1/4=1 x 4/1

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Step-by-step explanation:

divide 1 by 1 over 4, then using cross multiply

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3 years ago
Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The p
Ymorist [56]

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

Z = \frac{X - \mu}{s}

X = 7.1

Z = \frac{7.1 - 6.3}{0.3}

Z = 2.67

Z = 2.67 has a pvalue of 0.9962

X = 5.5

Z = \frac{5.5 - 6.3}{0.3}

Z = -2.67

Z = -2.67 has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

Z = \frac{6.7 - 6.3}{0.3}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

X = 5.9

Z = \frac{5.9 - 6.3}{0.3}

Z = -1.33

Z = -1.33 has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

5 0
3 years ago
HELP PLS YALL HDHDGDUSHSHXHHXXHHXHXBXBXXHDBBDBDBDBDD
Vedmedyk [2.9K]

Answer:

with what?

Step-by-step explanation:

5 0
3 years ago
Which line segments are diameters of circle U? choose all answers that are correct.
fomenos
B and D is correct



PS QT



GOOD LUCK
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3 years ago
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