Answer:
The probability that the mean of this sample is less than 16.1 ounces of beverage is 0.0537.
Step-by-step explanation:
We are given that the average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 ounces.
A random sample of sixty-five 16-ounce beverage cans are selected
Let = <u><em>sample mean amount of a beverage</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean amount of a beverage = 16.18 ounces
= standard deviation = 0.4 ounces
n = sample of 16-ounce beverage cans = 65
Now, the probability that the mean of this sample is less than 16.1 ounces of beverage is given by = P( < 16.1 ounces)
P( < 16.1 ounces) = P( < ) = P(Z < -1.61) = 1 - P(Z 1.61)
= 1 - 0.9463 = <u>0.0537</u>
The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9591.
Answer:
-x^8y^2 -2y^7 -2xy^4 -6xy
Step-by-step explanation:
Eliminate parentheses.
xy^4 + 5y^7 - 6xy - 7y^7 - x^8y^2 -3xy^4
Arrange in descending degree order, group like terms.
-x^8y^2 +(5y^7 -7y^7) +(xy^4 -3xy^4) -6xy
Combine like terms.
-x^8y^2 -2y^7 -2xy^4 -6xy
Answer:
5 presents
Step-by-step explanation:
40 presents in 2 hours
20 presents in 1 hour (divide by 2)
20 presents in 60 minutes ( 1 hour is 60 minutes )
5 presents in 15 minutes ( divide by 4)