16) Assuming that the number of shoppers who buy a packet of crackers is approximately normal with a mean of 20 and standard deviation of 4, what is the approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack

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 approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

This probability is 1 subtracted by the pvalue of Z when X = 28. So

has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1

i think this could be a possible answer (1 / 84) -1

good luck! : d