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3 years ago

The graph of f ′(x) is continuous and increasing with an x-intercept at x = 0.

1 answer:

3 0

Because $f'(x)$ is increasing, you know that $f''(x)>0$ , which means $f$ must be concave upward over its domain, so A is strue.

By the same fact above, you also know that D must be true.

Relative extrema occur for points where $f'(x)=0$ , and you know that the graph of $f'(x)$ crosses the x-axis at $x=0$ , so this is also true.

That leaves B. Why is it false? Inflection points occur at points where $f''(x)=0$ , where the sign of $f''(x)$ changes to either side of $x$ . But you know that $f''(x)>0$ is always true, so there is no such $x$ that makes $f''(x)=0$ .

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The amount y of your friend's allowance if the amount she receives is $77 more than the amount x you receive.

sergeinik [125]

The amount y is equals to x+77

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3 years ago

Hello need help with this thanks!

lina2011 [118]

The answer is 28
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2 years ago

Which equation is a direct variation function? O y=x O y = x + 2 O y=& Oy=4

LiRa [457]

The answer is B , y= x + 2

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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.

6 0

3 years ago

Jenny has $2.40 deducted from her monthly allowance every time she forgets to do a chore. If she forgets three times in a month,

NeX [460]

Answer:

D -$7.20

Step-by-step explanation:

If she takes (substance) $2.40 every times she forgets to do a chores you have to substance
but first you have to multiply because she forgets 3 times
2.40 × 3 is 7.20
but since she taking it away it would be negative

so your answer is -$7.20

I hope this helps

4 0

3 years ago