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4 years ago

Please view the screenshot below

Mathematics

1 answer:

Blizzard [7]4 years ago

7 0

Answer:

b = 363

Step-by-step explanation:

Sum of a geometric Progression is expressed as attached.

a = 3 = the first term

r = 3 = the common ratio

n = 5 = number of term

Sum of GP = a(r^n - 1)/ r-1

= 3(3^5 - 1)/ 3-1

= 3(234 - 1) / 2

= 3(242) / 2

= 726/2

= 363

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My common core lesson 7 3.1

Verizon [17]

What is the question?

5 0

3 years ago

Which inequalities are true? Check all that apply. Three-fifths < 0 Four-sixths greater-than one-half Nine-elevenths < 1 T

Kipish [7]

$\frac{3}{5} < 0$

$\frac{8}{16} > \frac{1}{2}\\\\$

Solution:

We have to find the inequalities that are true

Option 1

$\frac{3}{5} < 0\\\\0.6 < 0$

0.6 is not less than 0

Thus this inequality is not true

Option 2

$\frac{4}{6} > \frac{1}{2}\\\\0.667 > 0.5$

0.667 is greater than 0.5

Thus this inequality is true

Option 3

$\frac{9}{11} < 1\\\\0.818 < 1$

0.818 is less than 1

Thus this inequality is true

Option 4

$\frac{8}{16} > \frac{1}{2}\\\\$

0.5 = 0.5

Thus the above inequality is not true

6 0

3 years ago

Quadrilateral ABCD is similar to quadrilateral EFGH. Find the measure of side FG.

dezoksy [38]

59

47 - 11 = 36
23 + 36 = 59

I hope this helps. Brainliest appreciated.

7 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.

6 0

4 years ago

Hazel wants to mix 12 pounds of her special blend coffee that will cost her $12.50 per pound. She uses Columbian and Sumatra cof

k0ka [10]

Let x be the amount of Columbian coffee in the mixture and y the amount of Sumatra coffee in the mixture, then
x + y = 1 . . . (1)
9.25x + 14.25y = 12.50 . . . (2)

(1) x 9.25 => 9.25x + 9.25y = 9.25 . . . (3)

(2) - (3) => 5y = 3.25 => y = 3.25/5 = 0.65

From (1), x + 0.65 = 1 => x = 1 - 0.65 = 0.35

1 pound of the mixture contains 0.35 Columbian coffee and 0.65 Sumatra coffee
Therefore, 12 pounds of the mixture will contain 0.35 x 12 = 4.2 pounds of Columbian coffee and 0.65 x 12 = 7.8 pounds of Sumatra coffee

3 0

3 years ago