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posledela
3 years ago
13

A decimal number with two digits is between 4.3 and 4.8. It's less than 4.71 and greater than 4.49. The digit in the tenths plac

e is even. What is the number?
Mathematics
1 answer:
zubka84 [21]3 years ago
5 0
I think what may be confusing is the fact that they also list two three digit numbers with two decimal places each. Since the number is greater than 4.49 with only two digits, it can't be 4.5 since the tenths place is odd. Same goes for less than 4.71. So the only answer would be 4.6--two digits, even tenths place, greater than 4.49 and less than 4.71.
Just some logical thinking, I guess. I think they were trying to trick you on this one.
Good luck.
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Answer:

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Step-by-step explanation:

Given Equation:

                    2(x + 4)= -5 + 1

Solving the Equation for 'x':

                  2(x + 4)= -4

              2(x) + 2(4)= -4\\

                      2x+8=-4

Subtracting '8' both sides:

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                        2x=-12

Dividing by '2' both sides:

                         x=-12/2\\

                           x=-6

The Solution for the equation is:  x=-6

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3 years ago
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Answer:

No, there is not enough evidence at the 0.02 level to support the manager's claim.

Step-by-step explanation:

We are given that the company's promotional literature states that 68% of the chips fail in the first 1000 hours of their use. Also, a sample of 900 computer chips revealed that 66% of the chips fail in the first 1000 hours of their use.

And the quality control manager wants to test the claim that the actual percentage that fail is different from the stated percentage, i.e;

Null Hypothesis, H_0 : p = 0.68 {means that the actual percentage that fail is same as the stated percentage}

Alternate Hypothesis, H_1 : p  0.68 {means that the actual percentage that fail is different from the stated percentage}

The test statistics we will use here is;

     T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } } ~ N(0,1)

where, p = actual percentage of chip fail = 0.68

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           n = sample size = 900

So, Test statistics = \frac{0.66 -0.68}{\sqrt{\frac{0.66(1- 0.66)}{900} } }

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Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have sufficient evidence to accept null hypothesis.

Therefore, we conclude that the actual percentage that fail is same as the stated percentage and the manager's claim is not supported.

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Answer:

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Step-by-step explanation:

Explanation

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$$\begin{aligned}&(9 c+5)(9 c-5)=(9 c)^{2}-(5)^{2} \\&(9 c+5)(9 c-5)=81 c^{2}-25\end{aligned}$$

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Answer:

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