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3 years ago

Solve for y. -5(6+7y)= -5(6+y)

Mathematics

2 answers:

Keith_Richards [23]3 years ago

7 0

Simplify to -30-35y=-30-5y
Move all ys to the right and move constants left, -30y= 0
Divide by -30
y=0

fredd [130]3 years ago

4 0

Answer:

y = 0

Step-by-step explanation:

<em>Step 1:</em> Simplify both sides of the equation.

−5(6+7y)=−5(6+y)

(−5) (6) + (−5) (7y) = (−5) (6) + (−5) (y) (Distribute)

−30 + − 35y = −30 + −5y

−35y − 30 = −5y −30

<em>Step 2</em>: Add 5y to both sides.

−35y − 30 + 5y = −5y −30 + 5y

−30y −30 = −30

<em>Step 3</em>: Add 30 to both sides.

−30y − 30 + 30 = −30 + 30

−30y = 0

<em>Step 4</em>: Divide both sides by -30.

−30y /−30 = 0 /−30

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1. 3in =5ft, 6in=10ft, 9in=15ft, 12in=20ft, 18in=30ft, 21in=35ft (multiples of 3 for the top row, the bottom row is multiples of 5. when the second column appears 3x2 is 6 and 5x2 is 10, and keep going. whatever number you multiply 3 by is the same number you multiply 5 by.)

a) 30ft
b) Actual is 2.5ft and would be the bottom row. You would divide 5 by 2. Do the same for 3 and it is 1.5in.

2. Scale 2in = 4ft.
The scale dimensions of a room is 14x7
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The scale drawing is 7in. 3.5×2=7in Then 3.5×4=14ft

Final dimensions, 28ft x 14 ft.

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3 years ago

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3 years ago

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10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

In this problem, the parameters are given as follows:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$ .

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by $t^{\ast} = 2.4$ .

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

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2 years ago

What is the constant proportionality of 3 cans and 9 balls

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3 years ago

A company loses $780 as a result of shipping delay the 6 owners of the company must share the loss equally.

tekilochka [14]

A) Change in profit for each owner = -780/8

B) Change in profit for each owner = -$130

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Number of owners = 6

The 6 owners must share the loss equally

The change in profit for each owner will be the original profit less 780/6

That is -780/6

Evaluating -780/6

Change in profit for each owner = -$130

Learn more here: brainly.com/question/17858380

7 0

3 years ago

Solve for y. -5(6+7y)= -5(6+y)​

Solve for y. -5(6+7y)= -5(6+y)