All categories

3 years ago

Which expression is equivalent to 4-2/2^-3 a.-16 b.-8 c.8 d.16

Mathematics

1 answer:

algol133 years ago

6 0

Answer:

First, solve the top and bottom independently:

Top portion - 4-2 = 2

Bottom portion - A number to the negative power (2^-3) is equal to one over the positive power (1/2^3) which also equals 1/8

Combine both portions now: 2/(1/8)

To remove the fraction at the bottom portion, you can multiple both the top portion and bottom portions by 8 which resolves the bottom portion to be 1/8 * 8 = 1 AND the top portion to be 2 * 8 = 16.

Write the resulting equation with both top and bottom portions again and solve: 16/1 = 16

Step-by-step explanation:

You might be interested in

Jurado picked 40 bags of apples on Monday and sold them at his fruit stand for $3:45 each. The following week he picked and sold

tekilochka [14]

I'm pretty sure that would be $89.70!

4 0

3 years ago

in football game the quarterback was tackle behind the line of the scrimmage and lost 7 yards represent the loss of 7 yards as a

umka21 [38]

Seven I hope the answer helped!!

7 0

3 years ago

What is x - 5y = 13 and 9y - x = -17 using elimination method

Genrish500 [490]

X +(-5y)= 13
-x+ 9y =(-17)
0+4y=-4
4 4
y=1
x+-5(1)=13
x-5=13
+5 +5
 x=18
y=1

6 0

3 years ago

On the interval [x,7] , at which x-value is the average rate of change 56? ????????

Ainat [17]

Answer:

Step-by-step explanation:

i got it right on the test

3 0

3 years ago

Read 2 more answers

Suppose a set of test scores is appproximately bell-shaped, with a mean of 68 and a range of 54. (a) The minimum test score is a

liberstina [14]

Answer:

a.) Minimum test score = 41

Maximum test score = 95

b.)An approximate value of the standard deviation is 9.

Step-by-step explanation:

The given set of test scores is approximately bell shaped.

The mean of the test scores, $\mu$ = 68

The range of the data set = 54.

a.) i.) Minimum test score = $\mu - \frac{range}{2} = 68 - \frac{54}{2} = 68 - 27 = 41$

ii.) The maximum test score = $\mu + \frac{range}{2} = 68 + \frac{54}{2} = 68 + 27 = 95$

b.) From the 68-95-99.7 rule, also known as the Empirical rule, tell us that an approximation of the range is 3 standard deviations either way from the mean. So from this rule we can say that the minimum can also be found from $\mu - 3\sigma$ and the maximum can also be approximated as $\mu + 3\sigma$ .

From the above we can say that the range can be written as

$3\sigma - (-3\sigma) = range$

∴ $6\sigma = range$

∴ $\sigma = \frac{range }{6} = \frac{54}{6} = 9$

Therefore an approximate value of the standard deviation is 9.

7 0

3 years ago