Answer:
2.28% probability that a person selected at random will have an IQ of 110 or greater
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that a person selected at random will have an IQ of 110 or greater?
This is 1 subtracted by the pvalue of Z when X = 110. So
has a pvalue of 0.9772
1 - 0.9772 = 0.0228
2.28% probability that a person selected at random will have an IQ of 110 or greater
5×4=20 , 12×4=48
53÷4 = 13 then
x=13
Answer: 72 grams.
Step-by-step explanation:
Let be "s" the amount of sugar in grams that must be used for 252 grams of coconut to keep the same ratio of coconut to sugar.
Knowing that in the recipe for the coconut cookies, should be 420 grams of coconut and 120 grams of sugar, and you only have 252 grams of coconut, you can set up this proportion to find "s":
Now, you need to solve for "s":
Answer:
7,173
x = 7
Step-by-step explanation:
Within the question it gives two examples.
x = 0 corresponds to 2000
x = 1 corresponds to 2001
If you pay close attention the x-value is always the same as the last digits in the year. So with 2007 the last digit is 7, which then leads us to determine that
x = 7
f (x) = -327 (7) + 9462
f (x) = -2289 + 9462 = 7173
The sum of these numbers is an even number and it is prime.
Therefore it is not an odd number and is not a composite number.
