Answer:
What's the best estimate for 26% of 44
Step-by-step explanation:
If the average amount of meat eaten per person is 261 pounds per year then the average amount of meat eaten per day is 11 ounces.
Given that the average amount of meat eaten per person is 261 pounds per year.
We are required to find the average number of ounces of meat eaten per day.
There are 16 ounces in a pound. So,the average amount of meat eaten per person per year will be 261*16=4176 ounces per year.
Average number of ounces of meat eaten per day is there are 365 days in a year=4176/365=11.44 ounces per day.
Now by rounding it to the nearest ounce we will get 11 ounces.
Hence if the average amount of meat eaten per person is 261 pounds per year then the average amount of meat eaten per day is 11 ounces.
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Yes because whether you flip two negative integers around, the answer would still remain the same since its addition.
For example:
-9+-3 = -12 = -3+-9
Answer:
It takes the word processor 117 minutes to word process and spell check 27 pages.
Step-by-step explanation:
With the information provided, you can use a rule of three to find how long it would take to word process and spell check 27 pages given that it takes 26 minutes to word process and spell check 6 pages:
6 pages → 26 minutes
27 pages → x
x=(27*26)/6=117 minutes
According to this, the answer is that it takes the word processor 117 minutes to word process and spell check 27 pages.
<h2>
Answer with explanation:</h2>
According to the Binomial probability distribution ,
Let x be the binomial variable .
Then the probability of getting success in x trials , is given by :
, where n is the total number of trials or the sample size and p is the probability of getting success in each trial.
As per given , we have
n = 15
Let x be the number of defective components.
Probability of getting defective components = P = 0.03
The whole batch can be accepted if there are at most two defective components. .
The probability that the whole lot is accepted :
∴The probability that the whole lot is accepted = 0.99063
For sample size n= 2500
Expected value : 
The expected value = 75
Standard deviation : 
The standard deviation = 8.53
