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3 years ago

Need answer to number 2 with steps ASAP

Mathematics

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

1) 10 $x^{4}$ y² + 15x³y³

Step-by-step explanation:

5x³y(2xy + 3y²)

10 $x^{4}$ y² + 15x³y³ Distribute 5x³y to (2xy + 3y²)

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Hey can you please help me posted i picture of question

Harrizon [31]

The sum to this polynomial is the first answer choice "A"

5 0

3 years ago

Read 2 more answers

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann

mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

The probability mass function of a Geometric distribution is:

$P(Y=y)=(1-p)^{y}p;\ y=0,12...$

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

$P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844$

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

$=(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780$

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable Y as follows:

$\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445$

Compute the standard deviation of the random variable Y as follows:

$\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88$

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

= P (Y ≥ 3.325)

= P (Y ≥ 3)

= 1 - P (Y < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064$

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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3 years ago

Paul earned $72,000 last year. If the first $30,000 is taxed at 10% and income above that is taxed at 19%,

MArishka [77]

25%.......................................

8 0

2 years ago

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Sergio works at a bakery and needs to cover

Elan Coil [88]

Answer:

The surface area that Sergio needs to frost on each cake is 3,238 square centimeters

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The surface area of the cylindrical cake is equal to the lateral area plus the area of the top of cylinder (remember that the bottom of the cake does not need frosting)

$SA=\pi r^{2}+2\pi r h$

we have

$r=40.2/2=20.1\ cm$ ---> the radius is half the diameter

$h=15.6\ cm$

$\pi=3.14$

substitute the given values

$SA=(3.14)(20.1)^{2}+2(3.14)(20.1)(15.6)$

$SA=1,268.59+1,969.16\\SA=3,237.75\ cm^2$

Round up to the nearest whole number

$SA=3,238\ cm^2$

therefore

The surface area that Sergio needs to frost on each cake is 3,238 square centimeters

5 0

3 years ago

The depth y (in inches) of a lake after x years is represented by the equation y=0.2x+42. How much does the depth of the lake in

wlad13 [49]

42.8 in
Step-by-step explanation:
Step one:
given data
we are told that the expression for the depth of a lake after x years is given by

Required
The value of y when x= 4
Step two:
substitute the value of x in the expression to get y

After 4 years the depth will increase by 42.8 in

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2 years ago