She would need to buy 75% of the dress Bc 100%-25% is 75%
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> .
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
and standard deviation
, the zscore of a measure X is given by:

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:

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



has a pvalue of 0.0384
3.84% probability that it has a low birth weight
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.