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3 years ago

1. 3a+63b=

1 answer:

4 0

1. 3(a + 21b)
2. 2b(4b + 3)
3. 2a(b - y)
4. 4ab(2a - b)
5. 4x^2 - 5x + 1
6. 2ax + 2ay - bx - by
7. ax + x - ay - y
8. 2x^2 - 9x + 10
9. -ax + bx + 3ay - 3by

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Solve for c. 8 ge-2 = co! 2 C-12

hjlf

Answer:

-5

Step-by-step explanation:

8/3c-2=2/3c-12

first we'll do the LCM of both sides denominator 3 and 3=3

and multiply this with both side

3×8/3c - 2×3= 3×2/3c - 3×12

8c-6= 2c-36

8c-2c=-36+6

6c=-30

c=-30/6

c= -5

6 0

3 years ago

3. Round off to the nearest hundredth (2 decimal places). a) 3.752 = b) 0.506 = c) 3.966 = d) 1.838 = e) 30.063 = f) 404.759 = 1

scoundrel [369]

Answer:

A 3.8 B.30.1 C.4 D.1.83 E.30.1 F. 404.8

Step-by-step explanation:

Round all the way down

7 0

3 years ago

Fifty specimens of a new computer chip were tested for speed in a certain application, along with 50 specimens of chips with the

Murljashka [212]

Answer:

Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

Step-by-step explanation:

$H_{0}$ : Mean speed of the new chip is the same as the old chip

$H_{a}$ : Mean speed of the new chip is greater than the old chip

to calculate the z-statistic we can use the formula:

z= $\frac{M_{n}-M_{o}}{\sqrt{\frac{s_{n}^2}{N_{n}} +\frac{s_{o}^2}{N_{o}} } }$

if we put the numbers then

z= $\frac{495.6-481.2}{\sqrt{\frac{19.4^2}{65}} +\frac{14.3^2}{65} } }$ =4.8171

The probability of this z-statistic is < 0.001 Therefore Under 99% confidence level we can say that mean speed of the new chips is greater than that of the old chips

4 0

3 years ago

Can you please explain

hichkok12 [17]

The hole is x+4=0 because (x+4) is canceling out so the hole x=-4
Hopes this loves.

7 0

3 years ago

Find x. Question 1 options: 6 3 9 12

fredd [130]

Maybe it’s 3 but it actually depends on the equation . . .

6 0

3 years ago