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stellarik [79]
2 years ago
14

What is the value of a^3, if a =7

Mathematics
2 answers:
alekssr [168]2 years ago
8 0
Ok so remember
x^n means x times itself n times
example
x^2=x times x
x^4=x times x times x times x
so

a^3
a=7
a^3=a times itself 3 times therefor
a^3=a times a times a
subsitute 7 for a
7^3=7 times 7 times 7
7^3=343
Elis [28]2 years ago
7 0
If a = 7, then we can substitute 7 for a in the equation a^{3}.

7^{3} = 7 · 7 · 7 = 343
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Graph the line whose y -intercept is 5 and whose x -intercept is −7 .
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Look at the vertical line on the graph and find five
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8 0
3 years ago
one quart of strawberries weights about 2 pounds . About how many quarts of strawberries would weighs 1\4
storchak [24]
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7 0
3 years ago
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Give an indirect proof of the following statement
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3 0
3 years ago
A marketing research company desires to know the mean consumption of milk per week among people over age 32. A sample of 440 peo
Softa [21]

Answer:

The 98% confidence interval for the mean consumption of milk among people over age 32 is between 3.3 and 3.5 liters.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.98}{2} = 0.01

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.01 = 0.99, so Z = 2.327.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 2.327\frac{0.8}{\sqrt{440}} = 0.1

The lower end of the interval is the sample mean subtracted by M. So it is 3.4 - 0.1 = 3.3 liters

The upper end of the interval is the sample mean added to M. So it is 3.4 + 0.1 = 3.5 liters

The 98% confidence interval for the mean consumption of milk among people over age 32 is between 3.3 and 3.5 liters.

7 0
2 years ago
PLEASE HELP ME. I NEED HELP​
scoundrel [369]

#3). $124.16 will be in amount of deposit

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=》 b = $398.05

3 0
3 years ago
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