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2 years ago

The smallest of 6 consecutive integers is p, and the greatest is q. What is the value of the average of p and q in terms of p?

Mathematics

1 answer:

Paladinen [302]2 years ago

3 0

Answer:

Option A.

Step-by-step explanation:

It is given that the smallest of 6 consecutive integers is p. So, all numbers are

$p,p+1,p+2,p+3,p+4,p+5$

The greatest is q.

$q=p+5$

Now, the value of the average of p and q in terms of p is

$Average=\dfrac{p+q}{2}$

$Average=\dfrac{p+p+5}{2}$

$Average=\dfrac{2p+5}{2}$

$Average=\dfrac{2p}{2}+\dfrac{5}{2}$

$Average=p+2.5$

Therefore, the correct option is A.

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The paper feed on a copying machine has room for a stack of paper 4.0 cm high. If 10 sheets of paper are 0.08 cm thick, how many

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I don't know but I know it is going to be a lot

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2 years ago

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

(-6)-(-9) I need help plz

Grace [21]

Think of a number line

You have -6 and -9

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3 years ago

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In the figure above, the vertices of square

Olin [163]

(C) 6 + 3√3

Explanation:

Area of the square = 3

a X a = 3

a² = 3

a = √3

Therefore, QR, RS, SP, PQ = √3

ΔBAC ≅ ΔBQR

Therefore,

$\frac{BQ}{BA} = \frac{QR}{AC}$

$\frac{BQ}{BA} = \frac{\sqrt{3} }{BA}$

In ΔBAC, BA = AC = BC because the triangle is equilateral

So,

BQ = √3

So, BQ, QR, BR = √3 (equilateral triangle)

Let AP and SC be a

So, AQ and RC will be 2a

In ΔAPQ,

(AP)² + (QP)² = (AQ)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

Similarly, in ΔRSC

(SC)² + (RS)² = (RC)²

(a)² + (√3)² = (2a)²

a² + 3 = 4a²

3 = 3a²

a = 1

So, AP and SC = 1

and AQ and RC = 2 X 1 = 2

Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR

Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3

Perimeter = 6 + 3√3

Therefore, the perimeter of the triangle is 6 + 3√3

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2 years ago

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3 years ago

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