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3 years ago

I need to get all of them solved by tomorrow, and I have no idea where to start with any of them

Mathematics

1 answer:

Tems11 [23]3 years ago

6 0

Answer:

1- x=4,3

2- x= 2, 5/6

3- x= i dont know

4- b-

d- $2x\frac{3}{2}$

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2. Which of the following points lies in the solution set of the inequality y>/=3x+10?

Alex_Xolod [135]

Sorry if I am wrong I am not 100% sure but I think it’s 3

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2 years ago

A single die is rolled. Find the odds in favor of rolling a number greater than 5

NeX [460]

Answer:

A die has 6 numbers.

So 6 numbers them being
1, 2, 3, 4, 5, 6

There’s only one number greater then 5 which is 6.
Since its 6 numbers, a die has a 1/6 chance of getting your desired number.

So it’s a 1/6 chance.

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2 years ago

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ankoles [38]

A. A number line with an open circle on -4, shading to the left, and a closed circle on 3, shading to the right

8 0

3 years ago

Read 2 more answers

ADE

valina [46]

Answer:

7 cups of sugar

Step-by-step explanation:

Find out how many pies Luca need to bake left =

45 - 17 = 28

1 cup of sugar = 4 pies

28 ÷ 4 = 7 cups of sugar

This may be wrong as this was rushed as I have to go for my next class now.

4 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