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2 years ago

4.

Mathematics

2 answers:

alexandr1967 [171]2 years ago

7 0

Answer:

c.6xy and -16xy

Step-by-step explanation:

the variables are the same

yaroslaw [1]2 years ago

3 0

C is your answer because 6xy and -16xy have the same terms

You might be interested in

What is 0.04 written as a percent?

V125BC [204]

U can do this 2 ways....
(1) multiply it by 100.....0.04 * 100 = 4%
(2) or move the decimal two spaces to the right = 4%

3 0

3 years ago

The following gambling game has been proposed, which a player must pay to play. First, a value U is chosen uniformly from the se

Vesna [10]

Answer:

The player should be required to pay $5 to make this a fair game.

Step-by-step explanation:

U ~ Uniform(0, 10)

E[U] = (0 + 10)/2

= 5

X | U ~ Poisson(U)

E[X | U] = U

By law of total probability for expectations,

E[X] = E[E[X|U]] = E[U] = $5

Therefore the player should be required to pay $5 to make this a fair game.

6 0

3 years ago

Internet providers: In a survey of 935 homeowners with high-speed Internet, the average monthly cost of a high-speed Internet pl

BlackZzzverrR [31]

Answer:

68% of plans cost between $62.16 and $86.52.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean: 74.34

Standard deviation: 12.18

Bell-shaped is the same as normally distributed.

Estimate the number of plans that cost between $62.16 and $86.52.

62.16 is one standard deviation below the mean.

86.52 is one standard deviation above the mean.

By the Empirical rule, 68% of the measures are within 1 standard deviation of the mean.

68% of plans cost between $62.16 and $86.52.

7 0

2 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

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

7 0

2 years ago

Evaluate the expression (35+ 7) {6 – 5).

Fantom [35]

Answer:

The answer is 41

Step-by-step explanation:

(35 + 7) (6 - 5)

(42) × (1) = 41

Thus, The answer is 41

<u>-TheUnknownScientist</u>

8 0

2 years ago