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Gwar [14]
3 years ago
5

Solve -2(5x - 3) = 41. O x- 11.7 Ox-4.7 OX=-3.5 Ox-4,7

Mathematics
1 answer:
7nadin3 [17]3 years ago
7 0
Option b is the correct answer
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The y intercept is 1.5x+22.7 + the linear equation which is 22
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CAN SOMEBODY PLEASE HELP ME!!
VMariaS [17]

Answer:

2,2   2,7   5,7

Step-by-step explanation:

6 0
3 years ago
And can anyone do this?!!
Annette [7]

a. The interest earned will be $36.

b. The account balance will be $336.

Step-by-step Explanation:

Step 1:

The initial amount put into the account was $300.

This account has an interest rate of 4% every year. So we must calculate how much 4% of $300 is.

4% of $300 = \frac{4}{100} (300) = 0.04 (300) = 12.

Step 2:

So for every year, the interest of $12 is added into the account. To calculate total interest we multiply this interest amount and the number of years.

So after 3 years = 3(12) = 36.

So account balance after 3 years 300+36 = 336.

So the interest earned is $36 and the account balance is $336 after three years.

6 0
4 years ago
What is the measure of angle B?
butalik [34]

d. 128°

have a nice day!!

3 0
3 years ago
A production supervisor at a major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases the m
zheka24 [161]

Answer:

a. n= 47

b. n= 128

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the new catalyst, XA-100, increases the mean hourly yield of a chemical process, that is known to be μ=750 (pounds per hour) with the current process.

You need to calculate the sample size to estimate the population with determined error margins.

To do so, since you have no population information, only that it is approximately normal distributed, you'll use the Student t statistic to get the sample size.

The formula of the margin of error (d) is:

d= t_{n-1: 1-\alpha/2} * (\frac{S}{\sqrt{n} })

I've  cleared the sample size of the formula

n= (S*\frac{t_{n-1; 1-\alpha /2} }{d} )^{2}

You need a sample size for the t-Student value and a standard deviation, that's why the information of a pilot study with n=5 and S= 19.62 is given.

a)

95% CI

d= 8 pounds

t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.025}  = t_{4;0.975} =2.776

n= (19.62*\frac{2.776 }{8} )^{2}

n= 46.35 ≅ 47

b)

99% CI

d= 5 pounds

t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.005}  = t_{4;0.995} =4.604

n= (19.62*\frac{4.604}{8} )^{2}

n= 127.49 ≅ 128

I hope it helps!

4 0
3 years ago
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