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erma4kov [3.2K]
2 years ago
5

What percent is equivalent to 24/50? 24% 48% 50% 52%

Mathematics
2 answers:
stealth61 [152]2 years ago
8 0

Answer:

48%

Step-by-step explanation:

24 is almost 25 which is half of 50 you it would be 48%, hope this helps!:)

Dmitry [639]2 years ago
3 0
48 percent because 24/50 is almost 25 which would make it 50 put it’s not so it’s 48 have a great day
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The point-slope form:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (2, 2) and (0, -4). Substitute:

m=\dfrac{-4-2}{0-2}=\dfrac{-6}{-2}=3\\\\y-2=3(x-2)

The standard form:

Ax+By=C


y-2=3(x-2)        <em>use distributive property</em>

y-2=3x-6          <em>add 6 to both sides</em>

y+4=3x        <em>subtract y from both sides</em>

4=3x-y

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3 years ago
Which statement is correct? A) 13/ 3 &lt; 4.25 &lt; 43 /10 B) 43 /10 &lt; 13 /3 &lt; 4.25 C) 4.25 &lt; 13 /3 &lt; 43/ 10 D) 4.25
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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

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

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