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3 years ago

If f(x) = 7x^2 - 3x +1 and g(x) = 3x - 2, find the following by matching.

Mathematics

1 answer:

Cloud [144]3 years ago

4 0

f(x) = 7x² - 3x + 1

g(x) = 3x - 2

1. g(0) This means that x is 0, so you can plug in 0 for x in the equation:

g(x) = 3x - 2

g(0) = 3(0) - 2

g(0) = -2

2. g(1) x is 1

g(x) = 3x - 2

g(1) = 3(1) - 2

g(1) = 3 - 2

g(1) = 1

3. f(1) x is 1

f(x) = 7x²- 3x + 1

f(1) = 7(1)² - 3(1) + 1

f(1) = 7 - 3 + 1

f(1) = 5

4. f(x) = 7x²- 3x + 1

f(-2) = 7(-2)²- 3(-2) + 1

f(-2) = 7(4) + 6 + 1

f(-2) = 28 + 7

f(-2) = 35

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3 years ago

Exit Ticket Anyone Can help me?

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Answer:

okay, so 1 1/6 can be converted to 7/6.

2 3/8 can be converted to 19/8

now find a common denominator, which is 24

7/6 would turn into 28/24

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Step-by-step explanation:

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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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