I need the answer for part b

What is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D" id="TexFormula1" title="\frac{1}{5}" alt="\frac{1}{5}" align="a

gulaghasi [49]

the answer is 1

you just gotta use the forula for dividing fracitons which is keep the first fraction the sme do the opposite operation and flip the other fraction

8 0

3 years ago

What is the equation in point-slope form of the line that passes through the point (-5, 6) and has a slope of 2/3

Aleks [24]

Answer:

Step-by-step explanation:

The point-slope form of the equation of a line: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point the line passing through.

m = ²/₃

(-5, 6) ⇒ x₁ = -5, y₁ = 6

So the point-slope form of the equation:

y - 6 = ²/₃(x - (-5))

or: y - 6 = ²/₃(x + 5)

7 0

3 years ago

Round 4.2258..hours to the nearest minutes

RoseWind [281]

Answer:

253.548 minutes which rounds to 254 minutes

Step-by-step explanation:

1 hour = 60 minutes

4.2258 × 60 minutes = number of minutes per 4.2258 hours

253.548 minutes

The question does not specify how to round it so I will round to the nearest whole number: 254 minutes.

4 0

3 years ago

During the period of time that a local university takes phone-in registrations, calls come in

Zinaida [17]

Answer:

a) The expected number of calls in one hour is 30.

b) There is a 21.38% probability of three calls in five minutes.

c) There is an 8.2% probability of no calls in a five minute period.

Step-by-step explanation:

In problems that we only have the mean during a time period can be solved by the Poisson probability distribution.

Poisson probability distribution

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

a. What is the expected number of calls in one hour?

Calls come in at the rate of one each two minutes. There are 60 minutes in one hour. This means that the expected number of calls in one hour is 30.

b. What is the probability of three calls in five minutes?

Calls come in at the rate of one each two minutes. So in five minutes, 2.5 calls are expected, which means that $\mu = 2.5$ . We want to find P(X = 3).

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

$P(X = 3) = \frac{e^{-2.5}*(2.5)^{3}}{(3)!} = 0.2138$

There is a 21.38% probability of three calls in five minutes.

c. What is the probability of no calls in a five-minute period?

This is P(X = 0) with $\mu = 2.5$ .

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

$P(X = 0) = \frac{e^{-2.5}*(2.5)^{0}}{(0)!} = 0.0820$

There is an 8.2% probability of no calls in a five minute period.

6 0

3 years ago

Y=x-5 how do I graph<br>

Bezzdna [24]

Answer:

Draw a straight line through (0, -5) and (1, -4).

Step-by-step explanation:

Comparing the given y=x-5 to the general formula y = mx + b, we see that the slope of this line is m = 1 and that the y-intercept is b = -5.

Plot the y-intercept, (0, -5).

Now, starting with your pencil point on this (0, -5), move it 1 space to the right and 1 space up. This ensures a slope of m = 1/1, or just m = 1.

Your pencil point will now be at (1, -4).

Draw a straight line through (0, -5) and (1, -4).

4 0

4 years ago

I need the answer for part b​

I need the answer for part b