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3 years ago

What's the answer to a/4=1 1/2 ? please show work thanks

1 answer:

4 0

A/4=11/2 if you take 4/2 it equals 2 so you times the numerator and the denomerator by 2 and 11*2 equals 22 so the answer is a=22

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ASAAPPPPPP!!!!! Complete the tables of values

velikii [3]

Answer:

see explanation

Step-by-step explanation:

Using the rules of exponents

$a^{-m}$ ⇔ $\frac{1}{a^{m} }$ and $a^{0}$ = 1

Substitute the values of x into $4^{-x}$

x = 0 → $4^{0}$ = 1 ⇒ a = 1

x = 2 → $4^{-2}$ = $\frac{1}{4^{2} }$ = $\frac{1}{16}$ , hence

b = $\frac{1}{16}$

x = 4 → $4^{-4}$ = $\frac{1}{4^{4} }$ = $\frac{1}{256}$ , hence

c = $\frac{1}{256}$

6 0

3 years ago

Evaluate 13- (-9) need helpp

Viktor [21]

Answer:

The answer is 22.

Explanation:

Multiply -1 by -9.

$13 + 9$

Add 13 + 9.

$22$

Therefor, the answer is 22.

6 0

2 years ago

What is the solution of y= -5x + 1 and y= 3x - 2

Nezavi [6.7K]

Answer:

x=3/8, y=-7/8. (3/8, -7/8).

Step-by-step explanation:

y=-5x+1

y=3x-2

----------

-5x+1=3x-2

-5x-3x+1=-2

-8x+1=-2

-8x=-2-1

-8x=-3

8x=3

x=3/8

y=3(3/8)-2=9/8-2=9/8-16/8=-7/8

8 0

3 years ago

-39.85 + 34.8 - .375

Monica [59]

I believe it is -5.425

8 0

2 years ago

Read 2 more answers

Two distinct solid fuel propellants, type A and type B, are being considered for a space program activity. Burning rates of the

Finger [1]

Answer:

C.I. = (2.297, 11.703)

Step-by-step explanation:

The t-statistic for difference of mean is given by,

$t=\frac{\bar{x_{1}}-\bar{x_{2}}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

Here, $\bar{x_{1}}$ = 84

$\bar{x_{2}}$ = 7

s₁ = 4

n₁ = 12

s₂ = 6

n₂ = 18

Substituting all value in formula,

We get, t = -3.541 at 28 degree of freedom.

Using this formula, we get, t = 1.5342

Therefore, based on the data provided, the 99% confidence interval for the difference between the population means $\bar{x_{1}}-\bar{x_{2}}$ is: 2.297 < $\bar{x_{1}}-\bar{x_{2}}$ < 11.703

which indicates that we are 99% confident that the true difference between population means is contained by the interval (2.297, 11.703)

6 0

3 years ago