All categories

4 years ago

How do I slove this 9= m -4

Mathematics

2 answers:

alexandr402 [8]4 years ago

4 0

Answer:

m = 13

Step-by-step explanation:

m - 4 = 9

1.) Get rid of adding and subtracting.

m - 4 = 9

+ 4 + 4

m = 13

gladu [14]4 years ago

3 0

Hello there! I’d be happy to assist you with this question.

So we need to isolate M by itself to find out what it equals, you do that by the following:

You must add 4 to both sides in order to cancel out the -4 and isolate M.
9 = m - 4
+4 +4

Now - 4 + 4 = 0
9 + 4 = 13

So 13 = m

I hope this helped, and thank you for contributing to Brainly!

Regards,
-Luke

You might be interested in

Find tan 0 16 20 12 A. 19 B. D. 20

lisov135 [29]

Answer:

Answer is 16/12

Step-by-step explanation:

Tan is Opposite by Adjacent. the opposite side is 16, adjacent side is 12 and hypotenuse is 20.

so O/A= 16/12

8 0

3 years ago

Read 2 more answers

Find the value of 10! (10-3) A. 70 B. 120 O C. 5040 O D. 720

GarryVolchara [31]

Answer:

The answer is option D.

Step-by-step explanation:

$\frac{10!}{(10 - 3) ! } = \frac{10!}{7!} = \frac{3628800}{5040} = 720$

Hope this helps you

8 0

3 years ago

If a=7 and b+c=67, find a²+ab+ac

Stels [109]

Hope this will help..........

6 0

3 years ago

Read 2 more answers

Help me plz nowwwwww

Ymorist [56]

Answer:

...

Step-by-step explanation:

5 0

3 years ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n__1_-_n__2-2$

where, $\bar X_1$ = sample average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago