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5m - 7 = 6m + 11
Solve for M
5m - 6m = 11+ 7
-m = 18
Divide both sides by -1
-m/-1 = 18/-1
m = -18
Good luck!
Answer:
see the attachment
Step-by-step explanation:
<h2>
Answer with explanation:</h2>
Let
be the population mean lifetime of circulated $1 bills.
By considering the given information , we have :-

Since the alternative hypotheses is two tailed so the test is a two tailed test.
We assume that the lifetime of circulated $1 bills is normally distributed.
Given : Sample size : n=50 , which is greater than 30 .
It means the sample is large so we use z-test.
Sample mean : 
Standard deviation : 
Test statistic for population mean :-


The p-value= 
Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.
Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.
Answer:
41.88
Step-by-step explanation:
3.49*12=41.88
Answer:
Q1) (x+7)² = 9
x = -10, -4
Q2) (x-8)² = 144
x = -4, 20
Q3) (x-1)² = 81
x = -8, 10
Step-by-step explanation:
Q1) x² + 14x + 49 = 9
x² + 2(x)(7) + 7² = 9
(x + 7)² = 9
x + 7 = +/- sqrt(9)
x + 7 = 3
x = -4
x + 7 = -3
x = -10
Q2) x² - 16x + 64 = 144
x² - 2(x)(8) + 8² = 144
(x - 8)² = 144
x - 8 = +/- sqrt(144)
x - 8 = 12
x = 20
x - 8 = -12
x = -4
Q3) x² - 2x + 1 = 81
x² - 2(x)(1) + 1² = 81
(x - 1)² = 81
x - 1 =+/- sqrt(81)
x - 1 = 9
x = 10
x - 1 = -9
x = -8