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2 years ago

96 is 44 percent of what? Step by step

Mathematics

1 answer:

Elena-2011 [213]2 years ago

4 0

Answer:

$218.18$

Step-by-step explanation:

To find what number 96 is 44% of is we can use the formula:

$\frac{part}{whole} =\frac{percent}{100}$

Just substitute 94 and 44 into the equation:

$\frac{96}{x} =\frac{44}{100}$

$9600=44x\\218.1818...$

Or rounded,

$218.18$

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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti

melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that $m = 100, \mu = \frac{1}{100} = 0.01$

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

$P(X > x) = e^{-\mu x}$

This is P(X > 190). So

$P(X > 190) = e^{-0.01*190} = 0.1496$

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

$\ln{e^{-0.01x}} = \ln{0.08}$

$-0.01x = \ln{0.08}$

$x = -\frac{\ln{0.08}}{0.01}$

$x = 252.6$

Capacity of 252.6 cubic feet per second

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3 years ago

Amanda earned a score of 940 on a national achievement test that was normally distributed. The mean test score was 850 with a st

snow_lady [41]

<span>In math notation, we've done this: z = (X - μ) / σ = (940 - 850) / 100 = 0.90 where z is the z-score X is Vivian's score (940) µ is the mean (850) σ is the standard deviation (100) As you may know, in a normal distribution it's expected that about 68% of all observations will fall within 1 standard deviation of the mean, 95% will fall within 2 standard deviations, and 99% will fall within 3 standard deviations.
940 lie before the first standard deviation, in which 16.5% is above it
since 940 is 0.9 from the mean and 0.1 from the first standard deviation
so above it is 17.5 % = 0.175 or about 0.18 </span>

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