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Firdavs [7]
3 years ago
9

Which number has the greatest absolute value? A) 50 B) 30 C) −20 D) −60

Mathematics
1 answer:
DiKsa [7]3 years ago
6 0

Answer:

D -60

Step-by-step explanation:

-60 absolute value is 60 which is greater than the absolute value of 50, 30, and -20

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The volume of a sphere uses the following formula:

V = \frac{4}{3}\pi r^{3}

Because the question asks to use 3.14 in place of pi, the formula will now look like this:

V = (\frac{4}{3})(3.14)r^{3} = 4.18\overline{66} r^{3}

We are given a diameter of 118 feet. The radius is half of the diameter, so divide the diameter by 2 to find the radius:

118 \div 2 = 59
r = 59

Plug this value into the formula, and solve with a calculator:

4.18\overline{66} \times 59^3 = 4.18\overline{66} \times 205379 = 859853.413

Although our result is one cubic foot off, the answer is <span>A. 859,852ft^3.

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Solve the inequality. <br><br>The solution is.... ​
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Step-by-step explanation:

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4 years ago
HELPPPP PLEASEEEEE !!!!!What is the surface area of the composite solid?
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The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,
Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

If the mean number of arrivals is 10

This means that \mu = 10

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is P(X \leq 10)

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045

P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045

P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023

P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076

P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189

P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378

P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631

P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901

P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126

P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251

P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831

58.31% probability that there are no more than 10 arrivals.

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12+84
96
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