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2 years ago

Find the percent of increase. Original Price: $135 Retail Price: $162

Mathematics

1 answer:

dlinn [17]2 years ago

5 0

20%

Here's a tip: Always start percentage calculating with dividing the current number by 10.

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What’s the correct graph ?

maks197457 [2]

Answer:

The correct graph is graph E.

Step-by-step explanation:

We can tell this because we know the graph must stay below 3. So any graph over 3 gets thrown out. We also know that it must not go further down than -3. This is because any number over -3 as an input, when taking an absolute value, would be greater than 3.

Graph E shows numbers that are all less than 3 and greater than -3.

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3 years ago

Help with the pic down below

yulyashka [42]

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2 years ago

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2 years ago

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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.

6 0

2 years ago

Her jump rope Ella can make 17 jumps in 15 second.at this rate how make jumps can she make in20 minute?

shtirl [24]

Answer:

1360

Step-by-step explanation:

there are 1200 seconds in 20 minutes. if Ella can do 17 jumps in 15 seconds, you would divide 1200 by 15 (which equals 80) because that would be the number of 15 seconds in 20 minutes. then you would divide 80 by 17, which gives you your final answer of 1360 jumps in 20 minutes.

5 0

2 years ago