All categories

3 years ago

-8=-16+m pls show work! ty

Mathematics

2 answers:

puteri [66]3 years ago

7 0

Answer:

m = 8

Step-by-step explanation:

-8 = -16 + m

Switch sides.

-16 + m = -8

Add 16 to both sides.

m = 8

rosijanka [135]3 years ago

6 0

Answer:

-8=-16+m

-8+16=m

m=8

I would appreciate if my answer is chosen as a brainliest answer

You might be interested in

Given the two points what is the slope of the line? (4,2) (10.4)

zlopas [31]

$m = \frac{y2-y1}{x2-x1}\\ m=\frac{4-2}{10-4} \\m = \frac{2}{6} \\\\m=\frac{1}{3}$

Therefore, the slope of the line is 1/3. In case if you want the equation too.

$y=mx+b\\y=\frac{1}{3}x+b\\ 2=\frac{1}{3}(4)+b\\ 2=\frac{4}{3}+b\\ b=-\frac{4}{3}+2\\ b=-\frac{4}{3} +\frac{6}{3}\\ b=\frac{2}{3}$

Therefore, the equation is y=1/3x+2/3

6 0

3 years ago

Alan wants to bake blueberry muffins and bran muffins for the school bake sale. For a tray of blueberry muffins,

Novay_Z [31]

Answer:

A. 1/3x + 1/2y <= 4

2x + y <= 12

x >= 0

y >= 0

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Emma is making brownies. Her recipe says to use 3 eggs and 1 cup of vegetable oil. If she wants to make twice as many brownies,

erica [24]

Answer:

6 eggs and 2 cups of oil

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Anyone willing to do this

GalinKa [24]

Mmmmm no i just need pints

3 0

3 years ago

Read 2 more answers

For a quality control test, a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from it

amm1812

Answer:

95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

Step-by-step explanation:

We are given that a company checks the miles per gallon (mpg) for a simple random sample of 100 minivans drawn from its current inventory.

The average mpg of these 100 minivans is 15.6 with a standard deviation of 1.9 mpg.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average mpg = 15.6 mpg

s = sample standard deviation = 1.9 mpg

n = sample of minivans = 100

$\mu$ = population average mpg

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.987 < $t_9_9$ < 1.987) = 0.95 {As the critical value of t at 99 degree

of freedom are -1.987 & 1.987 with P = 2.5%}

P(-1.987 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.6-1.987 \times {\frac{1.9}{\sqrt{100} } }$ , $15.6+1.987 \times {\frac{1.9}{\sqrt{100} } }$ ]

= [15.22 mpg , 15.98 mpg]

Therefore, 95% confidence interval for the average mpg of all minivans in the company's inventory is [15.22 mpg , 15.98 mpg].

It is appropriate to compute a confidence interval for this problem using the Normal curve as t test statistics is used when data should follow normal distribution.

7 0

3 years ago