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lisabon 2012 [21]

3 years ago

12

Need help please, thank you

1 answer:

faust18 [17]3 years ago

5 0

M<ABX + m<CBX = 180

therefore, (14x + 70) + (20x + 8) = 180

Subtract 78 from both sides. Divide by 34.
x = 3

Plug in 3 for x.

m<ABX = 14(3) + 70 = 112
m<CBX =20(3) + 8 = 68

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IRINA_888 [86]

Answer:

24

Step-by-step explanation:

the number in center of both ranged

8 0

2 years ago

You are in the electronics store purchasing a video game that retails for 45$ the sales tax charged on the game is 2.70 what is

Katyanochek1 [597]

Answer:

6%

Step-by-step explanation:

let 't' = tax rate

45t = 2.7

t = 2.7/45

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2 years ago

The manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.The ma

Lana71 [14]

Answer:

The advertisement should use 16 minutes.

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 manager of a fast-food restaurant determines that the average time that her customers wait for service is 3.5 minutes.

This means that $m = 3.5, \mu = \frac{1}{3.5} = 0.2857$

What number of minutes should the advertisement use?

The values of x for which:

$P(X > x) = 0.01$

So

$e^{-\mu x} = 0.01$

$e^{-0.2857x} = 0.01$

$\ln{e^{-0.2857x}} = \ln{0.01}$

$-0.2857x = \ln{0.01}$

$x = -\frac{\ln{0.01}}{0.2857}$

$x = 16.12$

Rounding to the nearest number, the advertisement should use 16 minutes.

5 0

2 years ago

The auto parts department of an automotive dealership sends out a mean of 4.34.3 special orders daily. What is the probability t

Svetlanka [38]

Answer:

$P(X = 5) = 0.166$

Step-by-step explanation:

Given

$Mean = 4.3$

$x = 5$

Required

Determine the probability that the order is 5

This question can be answered using Poisson distribution.

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

Where

$\alpha$ is used to represent the mean

and

$\alpha = 4.3$

$x = 5$

<em /> $e = 2.71828$ <em> ---- Euler's constant</em>

So, we have:

$P(X = x) = \frac{\alpha ^x e^{-\alpha}}{x!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5!}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{5* 4 * 3 * 2 *1}$

$P(X = 5) = \frac{4.3^5 * 2.71828^{-4.3}}{120}$

$P(X = 5) = \frac{1470.08443 * 0.01356859825}{120}$

$P(X = 5) = \frac{19.9469850243}{120}$

$P(X = 5) = 0.1662248752$

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2 years ago

How long will it take you to drive 175 miles at a speed of 25 miles per hour?

BARSIC [14]

7 hours is the answer

7 0

3 years ago