Зу+9=14

Let X equal the number of typos on a printed page with a mean of 4 typos per page.

timama [110]

Answer:

a) There is a 98.17% probability that a randomly selected page has at least one typo on it.

b) There is a 9.16% probability that a randomly selected page has at most one typo on it.

Step-by-step explanation:

Since we only have the mean, we can solve this problem by a Poisson 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 time interval.

In this problem, we have that $\mu = 4$

(a) What is the probability that a randomly selected page has at least one typo on it?

Thats is $P(X \geq 1)$ . Either a number is greater or equal than 1, or it is lesser. The sum of the probabilities must be decimal 1. So:

$P(X < 1) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X < 1)$

In which

$P(X < 1) = P(X = 0)$ .

So

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X \geq 1) = 1 - P(X < 1) = 1 - 0.0183 = 0.9817$

There is a 98.17% probability that a randomly selected page has at least one typo on it.

(b) What is the probability that a randomly selected page has at most one typo on it?

This is $P = P(X = 0) + P(X = 1)$ . So:

$P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733$

$P = P(X = 0) + P(X = 1) = 0.0183 + 0.0733 = 0.0916$

There is a 9.16% probability that a randomly selected page has at most one typo on it.

3 0

3 years ago

Someone please help me go get BRAILIEST!!!!!!!!!

Oliga [24]

This answer would be letter C because if you simplify 49/56 divided by seven you get a grand total of 7/8.

5 0

3 years ago

HELp and show ur work Which of the expressions below has a product that is greater than 100? Select all that apply. A) 5.12 × 20

Bumek [7]

A & C should be your answer

6 0

3 years ago

Read 2 more answers

14 ounces of olive oil for $3.59

Kamila [148]

25. 6 cents for each ounce

5 0

4 years ago

What is 5x+538/23*1000-500=?

Leokris [45]

Answer:

x = (-105300)/23

Step-by-step explanation:

Solve for x:

5 x + 526500/23 = 0

Hint: | Put the fractions in 5 x + 526500/23 over a common denominator.

Put each term in 5 x + 526500/23 over the common denominator 23: 5 x + 526500/23 = (115 x)/23 + 526500/23:

(115 x)/23 + 526500/23 = 0

Hint: | Combine (115 x)/23 + 526500/23 into a single fraction.

(115 x)/23 + 526500/23 = (115 x + 526500)/23:

(115 x + 526500)/23 = 0

Hint: | Multiply both sides by a constant to simplify the equation.

Multiply both sides of (115 x + 526500)/23 = 0 by 23:

(23 (115 x + 526500))/23 = 23×0

Hint: | Cancel common terms in the numerator and denominator of (23 (115 x + 526500))/23.

(23 (115 x + 526500))/23 = 23/23×(115 x + 526500) = 115 x + 526500:

115 x + 526500 = 23×0

Hint: | Any number times zero is zero.

0×23 = 0:

115 x + 526500 = 0

Hint: | Isolate terms with x to the left hand side.

Subtract 526500 from both sides:

115 x + (526500 - 526500) = -526500

Hint: | Look for the difference of two identical terms.

526500 - 526500 = 0:

115 x = -526500

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 115 x = -526500 by 115:

(115 x)/115 = (-526500)/115

Hint: | Any nonzero number divided by itself is one.

115/115 = 1:

x = (-526500)/115

Hint: | In (-526500)/115, the numbers 526500 in the numerator and 115 in the denominator have gcd greater than one.

The gcd of 526500 and 115 is 5, so (-526500)/115 = (-(5×105300))/(5×23) = 5/5×(-105300)/23 = (-105300)/23:

Answer: x = (-105300)/23

5 0

4 years ago