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3 years ago

What is the probability that at least one of a pair of fair dice lands of 5, given that the sum of the dice is 8?

Mathematics

1 answer:

o-na [289]3 years ago

4 0

Answer:

0.40

Step-by-step explanation:

to find out the probability that at least one of a pair of fair dice lands of 5, given that the sum of the dice is 8

Let A = sum of dice is 8

B = one lands in 5

P(B/A) = P(AB)/P(A) by conditional probability

P(AB) = sum is 8 and one is 5

So (5,3) or (3,5)

P(A) = sum is 8.

i.e. (2,6) (2,6) (3,5) (5,3) (4,4)

Required probability

= n(AB)/n(A)

= $\frac{2}{5} =0.40$

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A dolphin swims 665,280 feet in 3 hours. Use the formula d = rt, where d represents distance, r represents rate, and t represent

lianna [129]

Answer:

Here we need to use the relation:

Distance = rate×time

Where "rate" is the speed at which the dolphin moves.

To answer different questions.

A) Dividing both sides by "time" we can rearrange the above relation to get:

rate = distance/time

B) We know that the dolphin swims 665,280 feet in 3 hours, then:

distance = 665,280 feet

time = 3 hours

Then the rate is

C) Now we want to rewrite the equation in miles per hour.

To do it, we need to use the relation:

1 mi = 5280 ft

Then we can rewrite the rate as:

That one is the rate in miles per hour.

D) Which one makes more sense to use?

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3 years ago

Simplify the expression. 2 (-5 - 7j)

Kaylis [27]

The answer is: -10-14j

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3 years ago

Read 2 more answers

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there

yan [13]

Answer:

Explained below.

Step-by-step explanation:

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The expected value of the sample mean of their weights is same as the population mean, μ = 1515 lbs.

The standard deviation of the sampling distribution of the sample mean weight is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{25}{\sqrt{16}}=6.25$

The average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 lbs. is:

$\text{Average Weight}=\frac{2500}{16}=156.25$

Compute the probability that a random sample of 16 persons on the elevator will exceed the weight limit as follows:

$P(\bar X > 156.25)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{156.25-151}{6.25})\\\\=P(Z>0.84)\\\\=1-P(Z$

4 0

3 years ago

Stanley wants to know how many students in his school enjoy watching talk shows on TV. He asks this question to all 24 students

wel

Answer:

The Answer is;

No, it is not a valid inference because his classmates do not make up a random sample of the students in the school.

Step-by-step explanation:

You cannot make a valid inference of the preference of watching TV talk shows by only considering students in one class. You have to randomly select people from the whole population (i.e the total number of students in the school) and then make an inference.

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3 years ago

A certain drink is made by adding 4 parts water to 2.5 part drink mix. What percent of the mixture is water? Omit % sign in your

Maru [420]

The mixture contains 61.5% of water.

Step-by-step explanation:

Total parts = 4+ 2.5 = 6.5

Parts of water =4

parts of drink = 2.5

Percent of water in the mixture = 4 × 100/ 6.5

= 400/6.5

= 61.53%

It is rounded to the nearest tenth as 61.5%.

8 0

3 years ago