A sandwich store charges a delivery fee to bring lunch to an office.
Delivery cost - D
Meatball sandwich cost - M
One office pays $40.50 for 5 meatball sandwiches.
40.50 = 5M + D
Another office pays $66.50 for 9 meat ball sandwiches
66.50 = 9M + D
A. How much does each meatball sandwich add to the cost of delivery?
D = 40.50 - 5M = 66.50 - 9M
40.50 - 5M = 66.50 - 9M
9M - 5M = 66.50 - 40.50
4M = 26.00
M = 6.50
B. What is the delivery fee
D = 40.50 - 5M = 40.50 - 5(6.50) = 40.50 - 32.50 = 8.00
Check
66.50 = 9M + D = 9(6.50) + 8 = 58.50 + 8.00 = 66.50
3 blocks east and 2 blocks north should be the correct answer :)
Answer:
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
Step-by-step explanation:
When the distribution is normal, we use 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 question, we have that:

The proportion of infants with birth weights between 125 oz and 140 oz is
This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So
X = 140



has a pvalue of 0.9772
X = 125



has a pvalue of 0.8413
0.9772 - 0.8413 = 0.1359
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
Answer:
the greatest common factor is 8
Let's do the math. In the example a =2, b = 3, c = 5. If we are trying to see if the sum of these numbers has a common factor more than 1 do the math. 2 + 3 + 5 = 10. Ten has four factors which are 1, 2, 5, and 10. So the answer to your question is yes.