50 you don't need to do anything else it already is approximate
Answer:
0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
A pretzel company calculated that there is a mean of 73.5 broken pretzels in each production run with a standard deviation of 5.1.
This means that
Find the probability that there will be fewer than 69 broken pretzels in a run.
This is the p-value of Z when X = 69.
has a p-value of 0.1894
0.1894 = 18.94% probability that there will be fewer than 69 broken pretzels in a run.
Answer:
#1= 5x5=25
25x the amount of squares= 25x6= 150cm^2
#2= 1.5x2.6= 3.9ft.
3.9xamount of right angle triangles= 3.9x8=31.2ft^2
(^2 means squared)
Answer:
The test results support the claim
Step-by-step explanation:
The Coca-Cola Company reported that the mean per capita annual sales of its beverages in the United States was 423 eight ounce servings (Coca-Cola Company website, February 3, 2009).
Suppose you are curious whether the consumption of Coca-Cola beverages is higher in Atlanta, Georgia, the location of Coca-Cola’s corporate headquarters.
A sample of 36 individuals from the Atlanta area showed a sample mean annual consumption of 460.4 eight ounce servings with a standard deviation of s = 101.9 ounces. Using α = .05, do the sample results support the conclusion that mean annual consumption of Coca-Cola beverage products is higher in Atlanta?
Hx: u <= 423 oz
Ha: u > 423 oz (claim)
z(460.4) = (460.4-423)/[101.9/sqrt(36)] = 2.2022
p-value = P(z > 2.2022) = 0.0138
Conclusion: Since the p-value is less than 5%
reject Hx at the 5% significance level.
The test results support the claim.