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3 years ago

Solve 2m-3-5= -m+10 PLEASE PLEASE PLEASE ANSWER

1 answer:

6 0

Solve for m:
2 m - 5 - 3 = 10 - m

Subtract like terms. -3 - 5 = -8:
2 m - 8 = 10 - m

Add m to both sides:
2 m + m - 8 = (m - m) + 10

m - m = 0:
2 m + m - 8 = 10

2 m + m = 3 m:
3 m - 8 = 10

Add 8 to both sides:
3 m + (8 - 8) = 8 + 10

8 - 8 = 0:
3 m = 10 + 8

10 + 8 = 18:
3 m = 18

Divide both sides of 3 m = 18 by 3:
(3 m)/3 = 18/3

3/3 = 1:
m = 18/3

The gcd of 18 and 3 is 3, so 18/3 = (3×6)/(3×1) = 3/3×6 = 6:
Answer: m = 6

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3 years ago

The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4. Suppose 64 golfers play

Vilka [71]

Answer:

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

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:

$\mu = 75, \sigma = 4, n = 64, s = \frac{4}{\sqrt{64}} = 0.5$

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

$Z = \frac{76 - 75}{0.5}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

6 0

3 years ago

Kristin spent $131 on shirts. Fancy shirts cost $28 and plain shirts cost $15. If she bought a total of 7 then how many of each

Tpy6a [65]

Hi there!!

First, Let plain shirts be x
Let fancy shirts be y
x + y = 7
15x + 28y = 131
Multiply first equation by 28 to eliminate y:
28x + 28y = 196
15x + 28y = 131
Subtracting the two equations:
13x = 65
x = 5 
Substituting x to get y:
x + y = 7
5 + y = 7
y = 2 
Answer: 5 plain shirts and 2 fancy shirts

Hope this helps u!!
Have an amazing rest of ur day:)))

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2 years ago

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tensa zangetsu [6.8K]

Answer:

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