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Shkiper50 [21]
2 years ago
7

provide a counterexample (specific values of a,b, etc. which make the statement false) for each of the following statements. Ass

ume that a,b,c,d ∊ ℤ for all statements. 1. If a I bc, then a I b or a I c. 2. If ab I cd , then a I b and c I d. 3. If a I b^2 , then a I b.
Mathematics
1 answer:
ivolga24 [154]2 years ago
8 0

Answer:

1. a=3, b = 2, c = 1.

2. a=2, b=3, c=4, d=6.

3. a = 9, b = 3.

Step-by-step explanation:

1.  3 | 21   is a counterexample  because 3 does not divide into  2 or 1.

2. 23 | 46 is a counterexample because 2 does not divide into 3.

3. 9 | 3^2 is a counterexample because 9 does not divide into 3.

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Answer:

Quantitative

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Step-by-step explanation:

The number of days traveled by randomly selected employees is a quantitative variable because it can be presented meaningful numerical form and further the number of days traveled by randomly selected employees are discrete because the number of days can be counted. Now, the level of measurement for the variable "number of days traveled by employees" is  ratio scale because it has a meaningful zero. Meaningful zero in this case will means that none of the day is spent on travelling.

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Rewrite the expression as a single exponent
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9^7

Step-by-step explanation:

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Marcos has 3/5 of a pizza remaining and gives his friend 1/5 of the remaining pizza. How much pizza is left over?
LuckyWell [14K]

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at first I said 1/2 but it's 2/5's.

Step-by-step explanation:

\frac{3}{5}  -  \frac{1}{5 }  =  \frac{2}{5}

3 0
3 years ago
GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m
Harrizon [31]

Answer:

(a) Null Hypothesis, H_0 : \mu \leq 3.50 mg  

     Alternate Hypothesis, H_A : \mu > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average mg of mercury in compact fluorescent bulbs.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, H_A : \mu > 3.50 mg     {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

                          T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean mg of mercury = 3.59

            \sigma = population standard deviation = 0.18 mg

            n = sample of bulbs = 25

So, <em><u>test statistics</u></em>  =  \frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }

                               =  2.50

The value of z test statistics is 2.50.

<u>Now, P-value of the test statistics is given by;</u>

         P-value = P(Z > 2.50) = 1 - P(Z \leq 2.50)

                                             = 1 - 0.9938 = 0.0062

<em />

<em>Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0
3 years ago
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