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3 years ago

Y/8=5 I don’t know where to start

Mathematics

2 answers:

Harrizon [31]3 years ago

7 0

Answer: so what you want to do is you have the equation.

Y/8=5

so you can just say it's Y divided by 8 = 5.

The inverse operation of division is multiplication.

so know it would be y×8=5. so you would multiply 8 by 5. which would be 40 so y=40. a way to check it is subsituting 40 for Y. 40/8=5

luda_lava [24]3 years ago

5 0

Answer: y = 40

Explanation: In this problem, notice that y is being divided by 8 so to

get y by itself, we need to multiply both sides of the equation by 4.

So we have (8)(y/8) = (5)(8).

Noice that on the left side, the 8 and 8

cancel each other out so we're just left with y.

On the right side, we have (5)(8) which is 40.

So we have y = 40.

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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

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

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

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

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.

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