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3 years ago

-7(-13+19) distributive property

Mathematics

1 answer:

Vanyuwa [196]3 years ago

6 0

Answer:

-42

Step-by-step explanation:

i put two documents for you. check them out and try to solve some math problems like this. hope it helps!:)

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Dvinal [7]

Answer: C

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3 years ago

HELP PLEASE!!

shutvik [7]

Answer:

6.5 x 10^6 To answer this question, you need to divide the mass of the sun by the mass of mercury. So 2.13525 x 10^30 / 3.285 x 10^23 = ? To do the division, divide the mantissas in the normal fashion 2.13525 / 3.285 = 0.65 And subtract the exponents. 30 - 23 = 7 So you get 0.65 x 10^7 Unless the mantissa is zero, the mantissa must be greater than or equal to 0 and less than 10. So multiply the mantissa by 10 and then subtract 1 from the exponent, giving 6.5 x 10^6 So the sun is 6.5 x 10^6 times as massive as mercury.

8 0

3 years ago

Read 2 more answers

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, 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 average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

2 years ago

Solve for w p=1.2w/h^2

andreyandreev [35.5K]

Solve for w:
p = (1.2 w)/h^2
(1.2 w)/h^2 = (6 w)/(5 h^2):
p = (6 w)/(5 h^2)
p = (6 w)/(5 h^2) is equivalent to (6 w)/(5 h^2) = p:
(6 w)/(5 h^2) = p
Multiply both sides by (5 h^2)/6:
Answer: w = (5 h^2 p)/6

6 0

3 years ago

charissa ordered 3 cans of lemonade for each person at her party. She also ordered 1 pizza for every 4 people. If she ordered 6

krek1111 [17]

Answer:

72 cans

Step-by-step explanation:

step 1

Find the number of people

we know that

She ordered 1 pizza for every 4 people

using proportion

Find out the number of people for 6 pizzas

$\frac{1}{4}=\frac{6}{x}\\\\x=24\ people$

step 2

Find the number of cans ordered

we know that

Charissa ordered 3 cans of lemonade for each person

using proportion

Find out the number of cans of lemonade she ordered for 24 people

$\frac{3}{1}=\frac{x}{24}\\\\x=24(3)\\\\x=72\ cans$

7 0

3 years ago