Which pair shows equivalent expressions?
A.2(2/5x + 2)=2 2/5x + 1
B.2(2/5x + 2)=4/5x + 4
C.2(2/5x + 4)=4/5x + 2
D.2(2/5x + 4)=2 2/5x + 8
Solution:
Let us distribute 2 inside the parenthesis.
That is, we use distributive property:
a(b+c)=ab+ac
So,
Answer:Option (b)
Applying distributive property, a(b+c)=ab+ac
So, Option (B) is correct.
Answer:
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?
This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So
X = 160
By the Central Limit Theorem
has a pvalue of 0.7257
X = 150
has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled