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OleMash [197]
3 years ago
8

Which expression is equivalent to 4÷147

Mathematics
1 answer:
mote1985 [20]3 years ago
7 0

Answer:

what are the choices???

Step-by-step explanation:

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It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

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}}.

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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0
3 years ago
Find the other endpoint of the line segment with the given endpoint and midpoint.
Marat540 [252]

Answer:

(-2,12)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Which of the following best describes the slope of the line below?
strojnjashka [21]

Answer:

the answer is D: undefined becous it's lies at Y-axis

3 0
3 years ago
Hello, could you help me with my math exercise? thank you in advance
Natali [406]

Answer:

100(x - 1)

Explanation:

Let's begin writing down everything

(X×(-11)+5)×(-9)+X-55

Now we can multiply

-9(-11X+5)+X-55

99X-45+X-55

And so we get

100X-100

Factorize 100

100(x-1)

4 0
3 years ago
Read 2 more answers
The product of two consecutive even integers is 360360. find the integers.
professor190 [17]
There are no 2 consecutive integers which give a product 360360
5 0
3 years ago
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