A consumer advocate group selects a random sample of 205 snow foods in find 174 of them contain more what is indicated on the snacks nutritional information. Create a 90% confidence interval for the population proportion for a snack foods that contain more fat grams and was indicated on the snacks nutritional information

(0.809, 0.891)

Step-by-step explanation:

Let p = population proportion for a snack foods that contain more fat grams and was indicated on the snacks nutritional information .

As per given , we have

Sample size : n= 205

Sample proportion:

z-value for 90% confidence = 1.645

Confidence interval for p:

Hence, the required confidence interval = (0.809, 0.891)

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.

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6