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german
3 years ago
8

Kristen was asked to write each of the numbers in the expression 80,000 x 25 using exponents. Her response was (8 x 10 to the po

wer of 3) x 5 to the power of 2. Was Kristen's response correct? Explain.
Mathematics
1 answer:
Gnom [1K]3 years ago
6 0

Kriste respone is wrong

<em><u>Solution:</u></em>

Given that,

Kristen was asked to write each of the numbers in the expression 80,000 x 25 using exponents

Her response was (8 x 10 to the power of 3) x 5 to the power of 2

No, the response of Kristen is wrong

Let us first understand about exponents

An exponent refers to the number of times a number is multiplied by itself

Given expression is:

80000 \times 25

Here, 80000 has 4 zeros

Therefore, it can be raised as 10 power 4

80000 = 8 \times 10^4

<em><u>Thus the expression becomes:</u></em>

80000 \times 25 = 8 \times 10^4 \times 25

Also, 25 can be written as 5 power 2

80000 \times 25 = 8 \times 10^4 \times 5^2

Thus given expression is written as (8 x 10 to the power 4) x 5 power to the power 2

Thus Kristen respone is wrong, because he has raised 10 to power 3

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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
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Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

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A increase, because less stores means more people going to the other ones, which means more money for the owners, which means if they raise the proces, people still need cloths sothey will still buy

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