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2 years ago

Simplify the expression. (r9/r3)4

Mathematics

2 answers:

eimsori [14]2 years ago

6 0

Answer:

12 is the answer

Step-by-step explanation:

first r/r will cancel r

9/3*4 is left

simplifying 9/3 gives 3

hence answer is 3*4=12

plz mark brainliest

slamgirl [31]2 years ago

5 0

Answer:

$9r \div 3r = (3r)4 = 12r$

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Rewrite the function by completing the square f(x)=x^2-14x+63

Veseljchak [2.6K]

Answer:

f(x) = (x - 7)² - 14

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Standard Form: ax² + bx + c = 0
Vertex Form: f(x) = a(bx - c)² + d
Completing the Square: (b/2)²

Step-by-step explanation:

Step 1: Define

f(x) = x² - 14x + 63

Step 2: Rewrite

Separate: f(x) = (x² - 14x) + 63
Complete the Square: f(x) = (x² - 14x + 49) + 63 - 49
Simplify: f(x) = (x - 7)² - 14

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3 years ago

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9a.
(-13.2) + 8.1 = -5.1

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2 years ago

How many times can 67 go into 487

Alborosie

487/67= 7.2686567
rounded= 7.269

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3 years ago

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Which point on the coordinate plane is closest to ______ ?

Vinvika [58]

Answer:

its point D hope this helps :)

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2 years ago

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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.

6 0

3 years ago