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2 years ago

What is the range & domain of the set

Mathematics

1 answer:

koban [17]2 years ago

3 0

This question is solved using range and domain concepts. Doing this, it is found that:

The domain is: {-15, -4, 3, 25, 32}
The range is: {-29, -10, 17, 21, 23}

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Domain: In a data-set, the domain is composed by the input values.
Range: In a data-set, the range is composed by the output values.

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In a set of ordered pairs, the first term in a pair represents the input and the second represents the output.
In this set, the inputs are: -4, 25, -15, 32, 3
The outputs are: 23, 17, -29, 21, -10

Thus:

The domain is: {-15, -4, 3, 25, 32}
The range is: {-29, -10, 17, 21, 23}

A similar question is given at brainly.com/question/18141895

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Explain why 1/12+ 1/12+ 1/12is the same as 1/4.

Zigmanuir [339]

Answer:

1/4 = 3/12 = 1/12+1/12+1/12

Step-by-step explanation:

Because 1/4 equals 3/12.

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3 years ago

Charlotte spent 1/2 of the day shopping at the mall. She spent 1/4 of this time trying on jeans. What fraction of the days did C

Nadya [2.5K]

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1/8

Step-by-step explanation:

1/2 * 1/4 = 1/8

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2 years ago

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What is the mean of the sum of positive single digit integers

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3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

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2 years ago

Use the binomial expression (p+q)n

zvonat [6]

Answer:

{0.16807, 0.36015, 0.3087, 0.1323, 0.02835, 0.00243}

Step-by-step explanation:

The expansion of (p+q)^n for n = 5 is ...

(p+q)^5 = p^5 +5·p^4·q +10·p^3·q^2 +10·p^2·q^3 +5·p·q^4 +q^5

When the probability p=0.3 and q = 1-p = 0.7 the terms of this series correspond to the probabilities of 5, 4, 3, 2, 1, and 0 favorable outcomes out of 5 trials.

For example, p^5 = 0.3^5 = 0.00243 is the probability of 5 favorable outcomes in 5 trials where the probability of each favorable outcome is 0.3.

___

The attachment shows the calculation of these numbers using a graphing calculator. It lists them in reverse order of the expansion of (p+q)^5 shown above, so that they are the probabilities of 0–5 favorable outcomes in the order 0–5.

3 0

3 years ago