Out of 5 cards in the bag ...

-- There are 2 even numbers . . . 2 and 4 .

-- There are 3 odd numbers . . . 1, 3, and 5 .

So EITHER time, the probability of pulling an even number is 2/5 ,
and the probability of pulling an odd number is 3/5 .

The probability of pulling an even number the first time
AND an odd number the second time is

(2/5) x (3/5) = 6/25 = <em>24%</em> .

3 0

3 years ago

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Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig

Anon25 [30]

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 3466, \sigma = 546$

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2500 - 3466}{546}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384

3.84% probability that it has a low birth weight

3 0

3 years ago

What is equivalent to 1/2+1/6

m_a_m_a [10]

Answer:

4/6 or 2/3.

Step-by-step explanation:

You need to multiply 1/2 by 3/3 so you can can the denominater having a 6.

3 0

2 years ago