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Katena32 [7]
3 years ago
9

Find two numbers whose difference is 102 and whose product is a minimum. Step 1 If two numbers have a difference of 102, and one

of them is x + 102, then the other is $$ Incorrect: Your answer is incorrect. x. Step 2 The product of two numbers x and x + 102 can be simplified to be x2 Correct: Your answer is correct. seenKey 2 + 102 Correct: Your answer is correct. seenKey 102 x. Step 3 If f(x) = x2 + 102x, then f '(x) = $$ Correct: Your answer is correct. 2x+102. Step 4 To minimize the product f(x) = x2 + 102x, we must solve 0 = f '(x) = 2x + 102, which means x = -51 Correct: Your answer is correct. seenKey -51 . Step 5 Since f ''(x) = 2 , there must be an absolute minimum at x = −51. Thus, the two numbers are as follows. (smaller number) (larger number)
Mathematics
1 answer:
NISA [10]3 years ago
4 0

Answer:

The two numbers would be -51 and 51

Step-by-step explanation:

To find these, first set the equation for the first number as x. You can then set the second number as x + 102. Now, find their product.

x(x + 102) = x^2 + 102x

Now, to find the minimum, find the value of x in the vertex of this equation.

-b/2a = -102/2(1) = -102/2 = -51

So we know -51 is the first number. Now we find the second using the prewritten equation.

x + 102 = -51 + 102 = 51

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Answer: (a)  14.5\%

(b) Restaurant B has a significantly lower percentage of orders that are not accurate.

Step-by-step explanation:

Confidence interval for population proportion is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Given: Significance level : \alpha: 1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

For Restaurant A , The proportion of orders not accurate : p=\dfrac{55}{295}\approx0.19

Then , the Confidence interval for population proportion will be :-

0.19\pm (1.96)\sqrt{\dfrac{0.19(1-0.19)}{295}}\\\\=0.19\pm0.045=(0.145,0.235)=(14.5\%,23.5\%)\\\\\text{i.e.}14.5\%

Also, 95​% confidence interval for the percentage of orders that are not accurate at Restaurant​ B:0.138

By comparing both the lower confidence limit and upper confidence limit of the interval for Restaurant B is lower than the lower confidence limit of t and the upper confidence limit of the interval for Restaurant A.

Therefore, Restaurant B has a significantly lower percentage of orders that are not accurate.

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3 years ago
Ayden is trying to lose weight to compete in a wrestling competition. In order to lose weight, he must burn more calories than h
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Penelope is going to the carnival to ride the rides. It costs $20 to get into the carnival and ride tickets are $0.50 each. Writ
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Answer:

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Step-by-step explanation:

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A quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain bra
tensa zangetsu [6.8K]

Answer:

We conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

Step-by-step explanation:

We are given that a quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates.

Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories.

<u><em>Let </em></u>\mu<u><em> = average calorie content of a 12-ounce can.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 120 calories     {means that the average calorie content of a 12-ounce can is less than or equal to 120 calories}

Alternate Hypothesis, H_A : \mu > 120 calories     {means that the average calorie content of a 12-ounce can is greater than 120 calories}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                         T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average calories per can = 124 calories

             s = sample standard deviation = 6 calories

            n = sample of cans = 10

So, <em><u>test statistics</u></em>  =  \frac{124-120}{\frac{6}{\sqrt{10} } }  ~ t_9

                               =  2.108

The value of t test statistics is 2.108.

<em>Now, at 0.05 significance level </em><em>the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test</em><em>. Since our test statistics is more than the critical values of t as 2.108 > 1.833, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

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3 years ago
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