What does (x+5)squared+10=10 equal??

Nikolas calculated the volume of the prism. His work is shown below. A triangular prism. The triangular base has a base of 5.8 f

Brut [27]

Answer:

B he used the incorrect formula for the volume of the prism

3 0

3 years ago

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Slope intercept form that passes though the point (-2,-3) and has a slope of 5/2

olganol [36]

$\text{Given that,}\\\\(x_1,y_1) = (-2,-3), ~~\text{and slope,}~~m = \dfrac 52\\\\\\\text{Now,}\\\\y -y_1 =m(x-x_1)\\\\\implies y +3 = \dfrac52 (x+2)\\\\\implies y = \dfrac 52(x+2) -3\\\\\\\implies y = \dfrac{5}2 x + 5 -3 \\\\\\\implies y = \dfrac 52 x + 2\\\\\\\text{This is the slope-intercept form(y =mx+b)}.$

8 0

3 years ago

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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:

$P(X > x) = 0.08$

So

$e^{-0.01x} = 0.08$

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

$-0.01x = \ln{0.08}$

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

$x = 252.6$

Capacity of 252.6 cubic feet per second

5 0

3 years ago

Find the quotient. Write your answer in simplest form.<br><br> 1/3 ÷ 6= 1/18?

Anna007 [38]

If you would like to solve 1/3 / 6, you can do this using the following steps:

1/3 / 6 = 1/3 / 6/1 = 1/3 * 1/6 = 1/18

The correct result would be 1/18.

8 0

3 years ago

Twelve notebooks cost $15.48 in all. What is the price of one notebook?

Ksju [112]

Answer:

$1.29

Step-by-step explanation:

Divide 15.48 by 12:

15.48 divided by 12= 1.29.

$1.29

7 0

3 years ago