The weights of bags of baby carrots are normally distributed, with a mean of 34 ounces and a standard deviation of .35 ounces. bags in the upper 4.5% are too ehavy and must be repackaged. what is the most a bag of baby carrots can weigh and does not need to be repackaged

The carrots can weight atmost 34.59 ounces so that it does not need to be repackaged.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 34 ounces

Standard Deviation, σ = 0.35 ounces

We are given that the distribution of weights of bags is a bell shaped distribution that is a normal distribution.

Formula:

We have to find the value of x such that the probability is 0.045

Calculation the value from standard normal z table, we have,  

Thus, the carrots can weight atmost 34.59 ounces so that it does not need to be repackaged.

3, 12, 48, 384, 1536, 6144, 24576, 98304

step-by-step explanation:

just keep timing it by 4.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.