Answer:
Step-by-step explanation:
um
We solve the inequality by subtracting 56.50 from both sides of the equation,
10.45b + 56.50 - 56.50 < 292.67 - 56.50
10.45b < 236.17
Then, divide both sides of the inequality by 10.45
b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22.
Answer:
84%
Step-by-step explanation:
The empirical rule tells you that 68% of the standard normal distribution is within 1 standard deviation of the mean. The distribution is symmetrical, so the amount in the lower tail is (1 -68%)/2 = 16%.
Since the number you're interested in, 240, is one standard deviation above the mean (200 +40), the percentage of interest is the sum of the area of the central part of the distribution along with the lower tail:
68% + 16% = 84%.
Answer:
70,110,110,110
Step-by-step explanation:
I'm Asian and I was less than 5 minutes
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
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.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:

What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So



has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79