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3 years ago

Find the volume of the cylinder 13mm and 16mm answer choices

Mathematics

1 answer:

Margaret [11]3 years ago

6 0

Answer:

Sorry about that guy

These are the answers still working January 9th of 2019

1.B

2.C

3.B

4.A

Late response but this will help future users at least :P

Step-by-step explanation:

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Sandy paid $23.40 for a handbag. The price of the handbag was $22.50. How much sales tax did she pay on the handbag?

xenn [34]

Answer:

90 cents

Step-by-step explanation:

23.40- 22.50= 0.90

5 0

3 years ago

Please help me it is timed and i am running outta time

Westkost [7]

Answer:

In ∆RST and ∆XYZ

RS=YZ(S)

angle RST=angle XYZ(A)

ST=YZ(S)

hence ∆RST ≠~∆XYZ

7 0

3 years ago

I really need this answer

otez555 [7]

Answer:

Step-by-step explanation:

If the lines are parallel, the slope of both of them are going to be the same. So if one line is 3, the other one will be too.

6 0

3 years ago

Read 2 more answers

A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular custome

MrRissso [65]

Answer:

- The probability that overbooking occurs means that all 8 non-regular customers arrived for the flight. Each of them has a 56% probability of arriving and they arrive independently so we get that

P(8 arrive) = (0.56)^8 = 0.00967

- Let's do part c before part b. For this, we want an exact booking, which means that exactly 7 of the 8 non-regular customers arrive for the flight. Suppose we align these 8 people in a row. Take the scenario that the 1st person didn't arrive and the remaining 7 did. That odds of that happening would be (1-.56)*(.56)^7.

Now take the scenario that the second person didn't arrive and the remaining 7 did. The odds would be

(0.56)(1-0.56)(0.56)^6 = (1-.56)*(.56)^7. You can run through every scenario that way and see that each time the odds are the same. There are a total of 8 different scenarios since we can choose 1 person (the non-arriver) from 8 people in eight different ways (combination).

So the overall probability of an exact booking would be [(1-.56)*(.56)^7] * 8 = 0.06079

- The probability that the flight has one or more empty seats is the same as the probability that the flight is NOT exactly booked NOR is it overbooked. Formally,

P(at least 1 empty seat) = 1 - P(-1 or 0 empty seats)

= 1 - P(overbooked) - P(exactly booked)

= 1 - 0.00967 - 0.06079

= 0.9295.

Note that, the chance of being both overbooked and exactly booked is zero, so we don't have to worry about that.

Hope that helps!

Have a great day :P

7 0

3 years ago

businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message

KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let X = number of text messages receive or send in an hour.

The random variable X follows a Poisson distribution with parameter λ.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: $\lambda=\frac{62.7}{24}= 2.6125$ .

The probability of a random variable can be computed using the formula:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...$

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

$P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180$

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667$

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0

3 years ago