All categories

2 years ago

Opening Exercise

Mathematics

1 answer:

Jobisdone [24]2 years ago

4 0

Answer:

100% of 180 = 180

25% of 180 = 45

-75% of 180 = -135

37.5% of 180 = 67.5

Step-by-step explanation:

I looked them up on an app so I hope its right!

You might be interested in

One of the plants in Jorge's garden grows at the rate of 2 Inches in 3 months. A second plant grows at the rate of 3 inches in 2

Molodets [167]

The second plant would because it grew one more inch in one less month (that's faster) than the other plant.

3 0

2 years ago

Give the domain and range.

3241004551 [841]

Answer:

The correct answer is a.

Step-by-step explanation:

The domain is the plotted points on the x-axis.

The range is the plotted points on the y-axis.

(-3, 3), (0, 0), (2, -2)

Domain {-3, 0, 2}; Range: {3, 0, -2}

8 0

2 years ago

Read 2 more answers

PLS HELPP

madam [21]

Use a calculator called symbolab.com 0-0

4 0

2 years ago

Ginny wants to compare the miles per gallon of 2 different cars. The cars mileage can be described as below. Find the miles per

solong [7]

Answer:

Car A gets 19 miles per gallon while Car B gets ~17.8 miles per gallon. Car A gets better mileage.

Step-by-step explanation:

3 0

2 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:

$P(X > x) = 0.08$

$e^{-0.01x} = 0.08$

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

$-0.01x = \ln{0.08}$

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

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

2 years ago