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yaroslaw [1]
3 years ago
11

Plz help will be thanked and will pick out brainly

Mathematics
1 answer:
vlabodo [156]3 years ago
8 0
B is your answer
Hope this helps:D
have a great rest of a brainly day!
You might be interested in
192x²y + 72.x” – 24rxy – 9px?
IRISSAK [1]

Answer:

3 ( 64 x 2 y + 24 x '' − 8 r x y − 3 p x )

Step-by-step explanation:

If you're looking to factor, which is what I'm guessing you're trying to do. This will be your answer ^^.

If you're looking to factor by grouping: 3 ( 64 x 2 y +24 x '' − 8 r x y − 3 p x )

Glad I could help!!

3 0
3 years ago
Simplify (8x4)(4x3)/7x2 A. 1/3x B. 1/28x C. 32x5/7 D. 7/9x4
umka21 [38]
C. 32x5/7 
is your answer

Hope this helps :)
4 0
3 years ago
Find a linear function f, given f(6) = - 4 and f(-3) = - 10. Then find f(0).
kiruha [24]

Answer:

y =  \frac{2}{3} x - 8

f(0)=-8

Step-by-step explanation:

Slope=rise/run=(-4-(-10))/(6-(-3))=6/9=2/3=m

y=m*x+q

-4=(2/3)*(6)+q

q=-8

f(0)=(2/3)*(0)-8=-8

6 0
4 years ago
Yashoda and Ingrid shared a sum of money in the ratio 3:2, respectively. Ingrid
Dmitry_Shevchenko [17]

Answer:

Yashoda got 1800

Step-by-step explanation:

Yashoda : Ingrid

3             :   2

Ingrid got 1200

Divide this by 2

1200/2 = 600

Multiply each side by 600

Yashoda : Ingrid

3*600      :   2*600

1800       : 1200

Yashoda got 1800

5 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
4 years ago
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