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Irina-Kira [14]
3 years ago
7

Computer Scored Unit Test

Mathematics
1 answer:
Zarrin [17]3 years ago
5 0
The correct and is bent which is a
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What is the greatest common factor of 72 and 36​
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The greatest common factor is 36
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0.21 divided by 33.2 !!! ASAP please show the work
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Answer:

0.0063253012

Step-by-step explanation:

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79=11x+13 <br> Solve each equation by using the properties of equality and justify each step
neonofarm [45]

Solution: X=6

Detailed explanation:

In this problem we're asked to find "x" or the unknown.

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---\mapsto\boldsymbol{-11x+79=13}

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---\mapsto\boldsymbol{-11x=-66}

To finish this off, divide both sides by -11.

---\mapsto\boldsymbol{x=6}

Voila! There's our answer!

<h2><em>Cheers!! ^-^</em></h2>

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Hope I helped.  Best wishes.

Reach far. Aim high. Dream big.

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8 0
2 years ago
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Justine paid 35.70 for a lego set that was on sale. The original price of the set was $42.00 what is the percentage discount for
Licemer1 [7]
15% because what you do is take $35.70 and subtract it from $42 then you get $6.3 and take that $6.3 and divide it from 42 and you get .15 but then just take the decimal away and it’s 15%.... don’t believe me just multiply 15% times 42
6 0
3 years ago
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Assume that females have pulse rates that are normally distributed with a mean of mu equals 72.0 beats per minute and a standard
larisa86 [58]

Answer:

a) 36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b) 66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c)

C.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

To solve this question, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 72, \sigma = 6.5

a. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 66 beats per minute and 78 beats per minute.

The probability is?

This is the pvalue of Z when X = 78 subtracted by the pvalue of Z when X = 66. So

X = 78

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

Z = \frac{78 - 72}{12.5}

Z = 0.48

Z = 0.48 has a pvalue of 0.6844

X = 66

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

Z = \frac{66 - 72}{12.5}

Z = -0.48

Z = -0.48 has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 66 beats per minute and 78 beats per minute

b. If 4 adult females are randomly​ selected, find the probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

The probability is?

Now we have n = 4, s = \frac{12.5}{\sqrt{4}} = 6.25

So

X = 78

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

By the Central Limit Theorem

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

Z = \frac{78 - 72}{6.25}

Z = 0.96

Z = 0.96 has a pvalue of 0.8315

X = 66

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

Z = \frac{66 - 72}{6.25}

Z = -0.96

Z = -0.96 has a pvalue of 0.1685

0.8315 - 0.1685 = 0.6630

66.30% probability that they have pulse rates with a mean between 66 beats per minute and 78 beats per minute

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The condition for the sample size exceeding 30 is when the population is skewed. If it is normally distributed, the size is not a condition.

So the correct answer is:

C.

Since the original population has a normal​ distribution, the distribution of sample means is a normal distribution for any sample size.

3 0
3 years ago
Read 2 more answers
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