Answer:
22.86% probability that the persons IQ is between 110 and 130
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:
If one person is randomly selected what is the probability that the persons IQ is between 110 and 130
This is the pvalue of Z when X = 130 subtracted by the pvalue of Z when X = 110.
X = 130
has a pvalue of 0.9772
X = 110
has a pvalue of 0.7486
0.9772 - 0.7486 = 0.2286
22.86% probability that the persons IQ is between 110 and 130
Answer:B the cost would be $11
Step-by-step explanation:
Solution:
1) Rewrite it in the form {a}^{2}-2ab+{b}^{2}, where a={d}^{2} and b=4
{({d}^{2})}^{2}-2({d}^{2})(4)+{4}^{2}
2) Use Square of Difference: {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}
{({d}^{2}-4)}^{2}
3) Rewrite {d}^{2}-4 in the form {a}^{2}-{b}^{2} , where a=d and b=2
{({d}^{2}-{2}^{2})}^{2}
4) Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)
{((d+2)(d-2))}^{2}
5) Use Multiplication Distributive Property: {(xy)}^{a}={x}^{a}{y}^{a}
{(d+2)}^{2}{(d-2)}^{2}
Done!
B) AAS Congruence Theorem
(correct answer i promise)
explanation:
The Angle Angle Side postulate (often abbreviated as AAS) states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.
