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3 years ago

Please help I have no idea how to do this and its due in a couple hours

Mathematics

1 answer:

stepan [7]3 years ago

3 0

Answer:for the first one, you should probably do a table of values then answer the question, i usually find that helpful

Step-by-step explanation:

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Find the probability that the hitter makes 10 outs in 10 consecutive at-bats, assuming at-bats are independent events. Hint:

emmainna [20.7K]

Answer:

0.037

Step-by-step explanation:

If the hitter makes an out 72% of the time, then the probabilty that he makes hit in 1 at-bat is $p=0.72$ and the probabilty that he doesn't make hit in 1 at-bat is $q=1-p=1-0.72=0.28.$

The probability that the hitter makes 10 outs in 10 consecutive at-bats, assuming at-bats are independent events is

$P=p^{10}=(0.72)^{10}\approx 0.037.$

5 0

3 years ago

What is 6.33 percent as a decimal

Harrizon [31]

I'm pretty sure the answer is .0633 but I'm not sure

5 0

4 years ago

Read 2 more answers

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

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

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0

2 years ago

A bookstore had 45 copies of a magazine. Yesterday, it sold 2/3 of them. Today, it sold 1/3 of what remained. How many copies do

Ket [755]

After the first day, 1/3 of the original amount remained:

... (1/3)·45 = 15

After the second day, 2/3 of that amount remained:

... (2/3)·15 = 10

The bookstore has 10 copies left.

6 0

4 years ago

How do i write 6/4 as a percent

babymother [125]

$\frac{6}{4} = 1.5$ × 100 = 150%

5 0

3 years ago