I believe it is 1, but i'm not completely sure.
Answer:
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
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:
What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So
X = 8.6
has a pvalue of 0.8413
X = 6.4
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
Let S = spoon
Let K = knife
1 knife is thee times the cost of a spoon, so it would be 3S
Now you have 9S + 12K = 82.80
We can replace the K with 3S:
9S +12(3S) = 82.80
Simplify the left side:
9S + 36S = 82.80
Add:
45S = 82.80
Divide both sides by 45"
S = 82.80 / 45
S = 1.84
One spoon cost $1.84
We know one knife cost 3 times as much, so multiply the cost of a spoon by 3:
Knife = 1.84 x 3 = $5.52
1/4 .....divide the pizza by 4 ...so you get 1/4 ...taken by each friend !