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3 years ago

Mrs. Hilt has 6 pizzas. Each pizza has 6 slices. She

Mathematics

2 answers:

Ede4ka [16]3 years ago

7 0

Answer:

Step-by-step explanation:

6 pizzas with six slices: 6*6=36---> she has 36 slices

1/4 of 36= how many she sold

1/4(36) or 36/4= 9

she sold 9 slices

36-9= the amount she has left

which is 27

Artyom0805 [142]3 years ago

4 0

Answer: 27 slices were left

Step-by-step explanation: If Mrs. Hilt had a total of 6 pizzas with 6 slices in each she would have 36 slices of pizza in total she sold 1/4 of the so set up you equation as 36/4 which equals 9 since it ask how many she had left you would subtract 9 from 36 {36-9=27)

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A computer rounded the number 129. 257 to the nearest hundredth what is this number rounded to the nearest hundredth

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3 years ago

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:

$P(X > x) = 0.08$