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vovikov84 [41]
3 years ago
9

Which choices are equivalent to the quotient below? Check all that apply √75/√15 A. 5 B. √25/√5 C. 15/3 D. √5 E. √3 F. √15/√3

Mathematics
2 answers:
Rudiy273 years ago
8 0
D. √5 <span> 
       
Let's first simplify the equation sqrt(75)/sqrt(15) Factor it (sqrt(3)sqrt(5)sqrt(5)) / (sqrt(3)sqrt(5)) Now cancel the sqrt(3) and sqrt(5) on both top and bottom, giving sqrt(5) / 1 sqrt(5) So we're just left with the square root of 5. Checking the available options, you see that D. </span><span>√5 </span> is an exact match. So that's your answer.
shusha [124]3 years ago
8 0

Answer:

The correct answers are B, D, and F

Step-by-step explanation:

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Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
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The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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 = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
3 years ago
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