10 cm 8 cm 2.5 cm 8 cm 6 cm 2.5 cm what’s the surface area

Karlie painted a portrait. The height of the portrait measures 14 inches. The width is half as long as the height. What is the a

NISA [10]

14 times 7, so 98 square inches

6 0

2 years ago

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What is the sum of the first five terms of a geometric series with a1 = 6 and r = 1/3? Express your answer as an improper fracti

Likurg_2 [28]

Answer:

242/27

Step-by-step explanation:

Using the formula for the sum, you can fill in the given values and do the arithmetic.

Sn = a1·(r^n -1)/(r -1)

S5 = 6·((1/3)^5 -1)/(1/3 -1) = 6(-242/243)/(-2/3) = 6·(121/81)

S5 = 242/27

_____

Or, you can write the terms and add them up. The sum is ...

6 + 2 + 2/3 + 2/9 + 2/27 = (162 +54 +18 +6 +2)/27 = 242/27

5 0

3 years ago

What is the area of a rhombus with dimensions of 22 inches and 10 inches?

docker41 [41]

The answer to the problem is 220

3 0

3 years ago

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Part A: If (72)x = 1, what is the value of x? Explain your answer. (5 points)

vichka [17]

Answer:

see below

Step-by-step explanation:

Part A: (72)^x = 1

Take the log base 72 of each side

log72(72^x) = log 72(1)

We know log a^b = b log a

x log72(72) = log72(1)

x = log72(1)

x = 0

Part A: (70)^x = 1

Take the log base 70 of each side

log70(70^x) = log70(1)

We know log a^b = b log a

x log70(70) = log70(1)

x = log70(1)

x = 0

7 0

3 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.

6 0

3 years ago