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madreJ [45]
3 years ago
8

(sum by gcf ) what is the gcf of 44+40? 44+40=4 (__+__)

Mathematics
2 answers:
kati45 [8]3 years ago
6 0
44|2
22|2
11|11
1

40|2
20|2
10|2
5|5
1

GCF(40,44)=2^2=4

44+40=4\cdot11+4\cdot10=4(11+10)
Illusion [34]3 years ago
4 0


The answer is 4

|________ you put 44+40 in the box

  |_______ then you see what goes into both and it is 4 so then divide 44 and 40 by 4

then put 11+10 in the box and nothing goes into 11 and 10 so 4 is the only thing that goes into 44 +40

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Vladimir79 [104]

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3 years ago
Elm Street is straight. Willard's house is at point H between the school at point S and
Vlad1618 [11]

Answer:

7.5 Miles.

Step-by-step explanation:

3 + 4.5 = 7.5 Miles.

4 0
3 years ago
motel clerk counts his $1 and $10 bills at the end of a day. He finds that he has a total of 57 bills having a combined monetary
madreJ [45]
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4 years ago
Find the terms and the coefficients of the expression.
Sav [38]
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3 0
3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

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.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7

This probability is 1 subtracted by the pvalue of Z when X = 204.1.

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

Z = \frac{204.1 - 201.9}{0.7}

Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0
3 years ago
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