Prove that AABE = ADCE, given that AE = DE, BE CE, with the same vertex measure ZE.
1 answer:
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Y =ax² + bx +c
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
Answer:
a) P(7.99 ≤ X ≤ 8.01) = 0.8164
b) P(X ≥ 8.01) = 0.0475.
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 can be 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
, the sample means with size n can be approximated to a normal distribution with mean
In this problem, we have that:
(a) Calculate P(7.99 ≤ X ≤ 8.01) when n = 16.
n = 16, so
This probability is the pvalue of Z when X = 8.01 subtracted by the pvalue of Z when X = 7.99. So
X = 8.01
Applying the Central Limit Theorem
has a pvalue of 0.9082
X = 7.99
has a pvalue of 0.0918
0.9082 - 0.0918 = 0.8164
P(7.99 ≤ X ≤ 8.01) = 0.8164
(b) How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X ≥ 8.01) =
n = 25, so
This is 1 subtracted by the pvalue of Z when X = 8.01. So
has a pvalue of 0.9525
1 - 0.9525 = 0.0475
P(X ≥ 8.01) = 0.0475.