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3 years ago

I need help!!! Please Help!!

Mathematics

1 answer:

stepladder [879]3 years ago

7 0

Answer:

Step-by-step explanation:

If you aren't supposed to do this on your calculator, then you'd have to figure out a way to get the bases of 10 to have the same power somehow in order to use the properties of exponents. This is the problem (you multiply the 2 dimensions together to find area, remember):

$(5.5*10^5)(4.2*10^4)$ You cannot simply add the exponents on the 10's and say your power is 9...cuz it's not. It needs to be rewritten so that there is a power of 4 on the 10 in the parenthesis on the left. Do that this way:

$(5.5*10^4*10^1)(4.2*10^4)$ Now you've got a common power of 4 between the 2 sets of parenthesis. 10 to the first is the same as 10, so multiplying that into the 5.5 gives us

$(55*10^4)(4.2*10^4)$

55 * 4.2 is 231 and 10 to the 4th times 10 to the 4th is 10 to the 8th.

$231*10^8$ But that's not in correct scientific notation. If we move the decimal to places to the left, we have to add 2 to the exponent of 8, giving us, finally,

$2.31*10^{10}m^2$

Next use the hint to convert that to kilometers:

$2.31*10^{10}m^2*\frac{1km^2}{1*10^6m^2}$

Dividing like bases means we subtract the lower exponent from the upper. 10 - 6 = 4, so the equivalent number of km squared is

2.31 × 10⁴ km²

Kilometers are comparable to miles, which is how we measure large things, like pieces of land. So it would be better to measure the forest in kilometers squared instead of meters squared.

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3 years ago

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

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3 years ago

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