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ss7ja [257]
3 years ago
10

Find the solution to the following system of equations using the addition method. 2x - y = 5 x + y = 13

Mathematics
2 answers:
leva [86]3 years ago
6 0

Answer:

x = 6

y = 7

Step-by-step explanation:

2x - y = 5

x + y = 13

2x - y = 5

y = 2x - 5

x + y = 13

x + 2x - 5 = 13

3x - 5 = 13

3x = 18

x = 6

y = 2x - 5

y = 2 ( 6 ) - 5

y = 7

zhenek [66]3 years ago
3 0

Answer:

Step-by-step explanation:

2x - y = 5

x + y = 13

3x = 18

x = 6

6 + y = 13

y = 7

(6,7)

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Of all the listeners who call the local radio station on the telephone , 60% are between the ages of 15 and 25. a random number
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3 years ago
The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a stan
IgorLugansk [536]

Answer:

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

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 sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 3.02, \sigma = 0.29, n = 20, s = \frac{0.29}{\sqrt{20}} = 0.0648

Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

This is 1 subtracted by the palue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 3.02}{0.0648}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907

1 - 0.8907 = 0.1093

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

5 0
4 years ago
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