You plug in the x = 3y+4 into the 3x part of 3x -5y=8. You distribute the 3 and add like terms to end up with 4y=12. You divide by 4 to get y alone and get -1. That’s your y value. You plug that -1 into the y part of any of the two equations to find x. The simplest one is to plug it in to x=3y+4, and you get -3+4=x, so x= -1
First, do the brackets (-14+12.6 = -1.4)
0.5 x -1.4 = -0.7
9.32+(-7.92) = 1.4
So…
0.7/1.4 = 0.5
So I believe the answer is 0.5
Answer:
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
Step-by-step explanation:
Problems of normally distributed samples can be 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:
Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when
The most that a bag can weigh and not need to be repackaged is 15.355 ounces.
There were 39 more adult tickets sold than children’s tickets