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3 years ago

Someone please help!!

Mathematics

2 answers:

Ivahew [28]3 years ago

7 0

G it would be g because it’s 13.5 in

just olya [345]3 years ago

6 0

Answer: I think the answer is F but I don’t know because I haven’t seen this in forever

Step-by-step explanation:

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Data on IQ for 33 first graders are given as follows: 82, 96, 99, 102, 103, 103, 106, 107, 108, 108, 108, 108, 109, 110, 110, 11

Dahasolnce [82]

Answer:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

Where n represent the sample size, on this case n =33. If we use this formula we got:

$\bar X = 113.727$

And for this case we know that is the best estimator since is an unbiased estimator for the true population mean since we have this:

$E(\bar X) = E(\frac{\sum_{i=1}^n x_i}{n})= \frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n} = \mu$

Step-by-step explanation:

For this case we have the following values given:

82, 96, 99, 102, 103, 103, 106, 107, 108, 108, 108, 108, 109, 110, 110, 111, 113, 113, 113, 113, 115, 115, 118, 118, 119, 121, 122, 122, 127, 132, 136, 140, 146

And we want to estimate the mean value of IQ for the conceptual population.

For this case we can use as estimator for the population mean the sample mean. We know that the sample mean is given by this formula:

$\bar X = \frac{\sum_{i=1}^n x_i}{n}$

Where n represent the sample size, on this case n =33. If we use this formula we got:

$\bar X = 113.727$

And for this case we know that is the best estimator since is an unbiased estimator for the true population mean since we have this:

$E(\bar X) = E(\frac{\sum_{i=1}^n x_i}{n})= \frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n} = \mu$

5 0

3 years ago

Point O is the center of the circle. Circle O is shown. Tangents D C and B C intersect at point C outside of the circle. Lines a

azamat

Answer:

The perimeter is 28 (6 plus 8=14 14x2= 28)

Step-by-step explanation:

Plz mark brainliest!

5 0

3 years ago

Read 2 more answers

The owner of a toy store received a shipment of 1,552 party favors. The party favors came in 28 boxes. The same number of party

Svetradugi [14.3K]

let

x = the number of party favors in each of the 27 boxes

y = the number of party favors in the last box

the total number of party favors is 1,552

27x + y = 1,552

since 1552 = 57*27 + 13

if we assume that y<x we conclude that there are 57 party favors in each of the 27 boxes and 13 party favors in the last box.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

8 0

3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago

Mr. Jacobs is going to make a histogram of the test scores from the last math test he gave. He plans to first organize the data

DaniilM [7]

Answer:

b 70-79

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers