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erastova [34]
3 years ago
7

WHATS IS -5x 2/7= I NEED HELP PLEASE

Mathematics
1 answer:
Kipish [7]3 years ago
4 0

-5x 2/7 =

−10x/7

srry if it is wrong man but that is what i got

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Find the measure of angle B
kirill [66]

Answer:

45 degree angle I believe

Step-by-step explanation:

Hope this helps!

Have a great day! :)

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2 years ago
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A store sells 12 cans of soup for $7.50 How much would it cost to purchase 6 cans of soup? $
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Answer: It would be $3.75

Step-by-step explanation: 12/2=6 and 7.50/2=3.75

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Simply this expression. Which like terms can she combine 12-4x+3
prohojiy [21]

like terms include terms that you can add/sub/mult/div - certain terms..

for example 23 and 7 are like terms, 4x and 56x are like terms....


here-> 12 and 3 are like terms so you can put them together:

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3 years ago
What is the area of the original rectangle?<br> A.8cm sq<br> B.20cm sq<br> C.25cm sq<br> D.40cm sq
Bumek [7]

Answer:

A. 8 cm sq

Step-by-step explanation:

Find the scale factor

new width/original width

8/1.6=5

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A=l×w

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A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of
notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.01 = 0.99, so Z = 2.327.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

M = z\frac{\sigma}{\sqrt{n}}

0.0001 = 2.327\frac{0.0003}{\sqrt{n}}

0.0001\sqrt{n} = 2.327*0.0003

\sqrt{n} = \frac{2.327*0.0003}{0.0001}

(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2

n = 48.73

Rounding up

49 measurements are needed.

7 0
2 years ago
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