Use technology or a z-score table to answer the question.

The number of baby carrots in a bag is normally distributed with a mean of 94 carrots and a standard deviation of 8.2 carrots.

Approximately what percent of the bags of baby carrots have between 90 and 100 carrots?

46%

Step-by-step explanation:

Since the number of baby carrots in a bag is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the number of baby carrots in the bag.

µ = mean

σ = standard deviation

From the information given,

µ = 94 carrots

σ = 8.2 carrots

The probability that a bag of baby carrots have between 90 and 100 carrots is expressed as

P(90 ≤ x ≤ 100)

For x = 90,

z = (90 - 94)/8.2 = - 0.49

Looking at the normal distribution table, the probability corresponding to the z score is 0.31

For x = 100,

z = (100 - 94)/8.2 = 0.73

Looking at the normal distribution table, the probability corresponding to the z score is 0.77

Therefore,

P(90 ≤ x ≤ 100) = 0.77 - 0.31 = 0.46

The percent of the bags of baby carrots that have between 90 and 100 carrots is

0.46 × 100 = 46%

step-by-step explanation:

consider the parallelogram shown in the attached image.

in the parallelogram hijk, the length of hk is given as,

from the properties of parallelogram we know that the opposite sides of a parallelogram are parallel to each other and are equal in measure.

therefore, the length of the side ij must be equal to the length of the side hk and that is .

hence, the expression for the length of ij confirms that hijk is a parallelogram.