Answer: -4860y^2
Solution:
81 x (5y x 6) x (-2y)
-81 x 5y x 6 x 2y
= -4860y^2
14.5 x 5.5x 0.5 = 39.875 cubic feet for the wall
brick = .5 x .333 x 0.167 = 0.0278 cubic feet rounded to 0.03
39.875/0.03 = 1329.16
so approximately 1330 bricks are required.
This could be a little different depending on how you round everything off
Answer:
637
Step-by-step explanation:
Ticket ending 00= £12 * 7 = £84
Tickets ending 5 = £1.5 * 75 = £112.5
£84+£112.5=£196.5
Price money + Profit: £196.5+£163= 356
356/£0.5 = 719 total tickets sold
719- 82 (WINNING TICKETS) = 637 losing tickets sold
Answer:
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
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:

The percentage of cockroaches weighing between 77 grams and 83 grams
This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So
X = 83



has a pvalue of 0.7734
X = 77



has a pvalue of 0.2266
0.7734 - 0.2266 = 0.5468
Rounded to the nearest whole number, 55%
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.