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3 years ago

20. The sum of two consecutive even integers is 158. Find the least of the two integers.

Mathematics

2 answers:

Aloiza [94]3 years ago

7 0

Answer:

2A + (2A +2) = 158

4A = 156

A= 39

2A = 78 and (2A + 2) = 80

Answer is A

Step-by-step explanation:

Mumz [18]3 years ago

6 0

Answer:

Step-by-step explanation:

The tricky part of this is figuring out how to assign the unknowns. We are told that we are working with two consecutive even integers. Consecutive means "next to" or "in order" and sum means to add. If we use 2 and 4 as examples of our 2 consecutive even integers and assign x to 2, then in order to get from 2 to 4 we have to add 2. So the lesser of the 2 integers is x, and the next one in order will be x + 2. (2 and 4 are just used as examples; they mean nothing to the solving of this particular problem. You could pick any 2 even consecutive integers and find the same rule applies. All we are doing here with the example numbers is finding a rule for our integers.) Now we have the 2 expressions for the integers, we will add them together and set the sum equal to 158:

x + (x + 2) = 158

The parenthesis are unnecessary since we are adding, so when we combine like terms we get

2x + 2 = 158 and

2x = 156 and

x = 78

That means that the lesser of the 2 integers in 78, and the next one in order would be 80, and 78 + 80 = 158

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f) 1

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

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$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

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

$Z = \frac{5.8-6}{0.2}$