Solve 3m+7/2=−2m+5/2 what is m

2 answers:

5 0

Solve for m:3 m + 7/2 = 5/2 - 2 m
Put each term in 3 m + 7/2 over the common denominator 2: 3 m + 7/2 = (6 m)/2 + 7/2:(6 m)/2 + 7/2 = 5/2 - 2 m
(6 m)/2 + 7/2 = (6 m + 7)/2:(6 m + 7)/2 = 5/2 - 2 m
Put each term in 5/2 - 2 m over the common denominator 2: 5/2 - 2 m = 5/2 - (4 m)/2:(6 m + 7)/2 = 5/2 - (4 m)/2
5/2 - (4 m)/2 = (5 - 4 m)/2:(6 m + 7)/2 = (5 - 4 m)/2
Multiply both sides by 2:6 m + 7 = 5 - 4 m
Add 4 m to both sides:6 m + 4 m + 7 = (4 m - 4 m) + 5
4 m - 4 m = 0:6 m + 4 m + 7 = 5
6 m + 4 m = 10 m:10 m + 7 = 5
Subtract 7 from both sides:10 m + (7 - 7) = 5 - 7
7 - 7 = 0:10 m = 5 - 7
5 - 7 = -2:10 m = -2
Divide both sides of 10 m = -2 by 10:(10 m)/10 = (-2)/10
10/10 = 1:m = (-2)/10
The gcd of -2 and 10 is 2, so (-2)/10 = (2 (-1))/(2×5) = 2/2×(-1)/5 = (-1)/5:Answer: m = (-1)/5

Dovator [93]3 years ago

4 0

Answer:

-2/5

Step-by-step explanation:

3m + 7/2 = -2m + 5/2

3m + 7 = -2m + 5

5m = -2

m = -2/5

You might be interested in

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.895 g and a standard deviation

kvasek [131]

Answer:

3.67% probability of randomly seleting 37 cigarettes with a mean of 0.809 g or less.

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 $s = \frac{\sigma}{\sqrt{n}}$ .