What is the square root of 34 to the nearest 0.05

Will give brainliest no files plz

jok3333 [9.3K]

Answer:

x=3 and y= 4 / 3 (Fraction)

Step-by-step explanation:

Substitute x for 6y-5.

2(6y−5)−3y=2

Add 6y and 3y, and multiply 5 by 2/Simplify the equation.

9y−10=2

Add 10 to both sides, and then divide each side by 9.

9y−10+10=2+10

9y / 9 = 12 / 9

y = 4/3.

Repeat the process but replace y with 4/3!

8 0

2 years ago

Read 2 more answers

Bryan owes his friend more than $1,930 from a loan. Bryan has already paid his friend back $200 and continues to pay him $173 ea

Ann [662]

Answer:

Choice C

Step-by-step explanation:

Bryan owes MORE than $1930 which means how much he pays back is going to be greater than 1930$

And so, $200 + $173 for certain number of months is going to be more than 1930.

Therefore, $200+ $173x >$1930

Hope this helped..

Have a great day : )

8 0

2 years ago

Mr Huey gas 54 sheets of paper he gave each boy 9 sheets.how many boys are there

emmainna [20.7K]

Answer:

There are six boys

Step-by-step explanation:

Divide 54 sheets by number of boys (9)

4 0

3 years ago

Read 2 more answers

(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

Write the sum as a product of two factors.

Elena-2011 [213]

A) 81 and 48 have a common factor of 3:
81w+48= 3(27w+16)

b) 10 and 25 have a common factor of 5:
10-25t=5(2-5t)

c) 12, 16 and 8 have common factors of 4:
12a+16b+8=4(3a+4b+2)

Hope this helps :)

4 0

2 years ago