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anzhelika [568]

3 years ago

9

Xiao's teacher asked him to rewrite the sum 60+ 90 as the product of the GFC of the two numbers and a sum. Xiao wrote 3(20+30).

What mistake did Xiao make? How should he have written the sum?

1 answer:

Dennis_Churaev [7]3 years ago

3 0

The greatest common factor here is not 3, it's 30.
so it should be 30(2+3)

hopefully this helps!

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What is the product of 3x(x + 4)<br> cOf

Lyrx [107]

Answer:

3x²+12x

Step-by-step explanation:

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6 0

3 years ago

Two plus ten times eight

alexdok [17]

Two plus ten times eight
(2+10)*8 = 96

3 0

3 years ago

F(x) = -4x2 + 4x – 4<br> Find f(3)

ValentinkaMS [17]

F(3) occurs when x =3. So lets plug 3 into our equation.

$f(3) = -4(3)^2 + 4(3)-4\\f(3) = -4(9) + 12- 4\\f(3) = -36 + 12- 4\\f(3)= -28$

F(3) = -28

I hope this helps! :)

5 0

2 years ago

Am I correct <br> 2017 -1846= 171<br> 2017-1781= 236<br> 1847-1781=65 yrs difference

Anit [1.1K]

Yes for the first one, yes for the second one, and yes for the third one although your subtraction equation had the wrong year (its not 1847 its 1846) but the answer was right, 65.

5 0

3 years ago

(b) how many measurements must be averaged to get a margin of error of ±0.0001 with 98% confidence? (round your answer up to the

goblinko [34]

Given that the scale readings of the laboratory scale is normally distributed with an unknown mean and a standard deviation of 0.0002 grams.

Part A:

If the weight is weighted 5 times and the mean weight is 10.0023 grams, the 98% confidence interval for μ, the true mean of the scale readings is given by:

$98\% \ C. \ I.=\bar{x}\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right) \\ \\ =10.0023\pm2.33\left(\frac{0.0002}{\sqrt{5}}\right) \\ \\ =10.0023\pm2.33(0.000089) \\ \\ =10.0023\pm0.00021 \\ \\ =(10.0023-0.00021, \ 10.0023+0.00021) \\ \\ =(10.0021, \ 10.0025)$

Thus, we are 98% confidence that the true mean of the scale readings is between 10.0021 and 10.0025.

Part B:

To get a margin of error of +/- 0.0001 with a 98% confidence, then

$\pm z_{\alpha/2}\left(\frac{\sigma}{\sqrt{n}}\right)=\pm0.0001 \\ \\ \Rightarrow2.33\left(\frac{0.0002}{\sqrt{n}}\right)=0.0001 \\ \\ \Rightarrow\left(\frac{0.0002}{\sqrt{n}}\right)= \frac{0.0001}{2.33} =0.0000429 \\ \\ \Rightarrow\sqrt{n}= \frac{0.0002}{0.0000429} =4.66 \\ \\ \Rightarrow n=4.66^2=21.7156\approx22$

Therefore, the number of <span>measurements that must be averaged to get a margin of error of ±0.0001 with 98% confidence is 22.</span>

8 0

3 years ago