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Dmitry_Shevchenko [17]
3 years ago
14

Perform the indicated operation (-25) - (25)

Mathematics
1 answer:
slavikrds [6]3 years ago
4 0

Hello,

(-25) - (25) = (-25) + (-25) = (-50)

Bye :)

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PLZZ I NEED THIS FAST!
Burka [1]

So remember that GCFs are greatest common factors, or the greatest number and/or variable that divides 2 or more terms nicely (aka no decimals).

With this, you at least know that a factor of these two numbers is 7, which turns this expression to 7(4 + 10). However, what's inside the parentheses can still be factored out with a GCF of 2. So you can further factor the expression as 2 * 7(2 + 5), which can be simplified to 14(2 + 5), or D. Since 2 and 5 do not have GCFs greater than 1, that is your answer.

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QT 1.
Delvig [45]
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6 0
3 years ago
62 is 90.5% of what number?
Reptile [31]
68.5. You take 62 and divide it by the percentage as a decimal, .905. 62 / .905 = 68.5.
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The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have 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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
Jill measures some scrap boards in her garage to
Alika [10]

Answer: Yes, Jill has enough scrap boards to create a border around her garden.

Step-by-step explanation:

2.75 + 3.2 + 1.65 + 2.6 = 10.2 m

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