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Phoenix [80]
3 years ago
8

If I multiply all whole numbers from 1 through 10 what is the largest power of 4 that is a factor of the product?

Mathematics
1 answer:
Oksanka [162]3 years ago
8 0
If we multiply 1 through 10, the mathematical representation of the product would be,
                        P = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10
Getting the factors of each number,
                         P = 1 x 2 x 3 x (2 x 2) x 5 x (3 x 2) x 7 x (2 x 2 x 2) x 9 x (5 x 2)
Then, we take out all the 2's.
                            2 x (2 x 2) x (3 x 2) x (2 x 2 x 2) x (5 x 2)
Combining all the 2's.
                             2^8 x 3 x 5
  2^8 can also be written as (2 x 2)^4. Thus, the answer is 4. 
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tom paid $16 for 6 audio cassettes. Which proportion can be used to find the amount of money (m) he would pay for 9 audio casset
tigry1 [53]

Answer:

Step-by-step explanation:

From the question, Tom paid $16 for 6 audio cassettes and we are told to find the proportion to get the amount of money (m) he would pay for 9 audio cassettes. To get this, we have to know the amount of the cost of 1 audio cassette. Since Tom paid $16 for 6 audio cassettes, 1 audio cassette will cost:

= $16/6

= $2.67

1 audio cassette will cost $2.67

The amount of money spent will depend on the number of audio cassettes. Therefore,

m = $2.67 × n

m = $2.67n

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You measure 44 textbooks' weights, and find they have a mean weight of 51 ounces. Assume the population standard deviation is 11
elena-14-01-66 [18.8K]

Answer: (47.51, 54.49)

Step-by-step explanation:

Confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

, where n= sample size .

\sigma = population standard deviation.

\overline{x} = sample mean

z_{\alpha/2} = Two -tailed z-value for {\alpha (significance level)

As per given , we have

\sigma=11.8\text{ ounces}

\overline{x}=51 \text{ ounces}

n= 44

Significance level for 95% confidence = \alpha=1-0.95=0.05

Using z-value table ,

Two-tailed Critical z-value : z_{\alpha/2}=z_{0.025}=1.96

Now, the 95% confidence interval for the true population mean textbook weight will be :-

51\pm (1.96)\dfrac{11.8}{\sqrt{44}}\\\\=51\pm(1.96)(1.7789)\\\\=51\pm3.486644\approx51\pm3.49\\\\=(51-3.49,\ 51+3.49)\\\\=(47.51,\ 54.49)

Hence, the 95% confidence interval for the true population mean textbook weight. :  (47.51, 54.49)

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