A would be (-3, 6).
B would be (-6, 3).
Anytime something is asking you to reflect off of the Y-axis, you put the original but flipped. Also, your answer will always be straight across vertically.
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
The equation should be
j = 30 + 3a
j + a = 210
Solving the equation
j + a = 210
(30 + 3a) + a = 210
30 + 4a = 210
4a = 180
a = 45
j = 30 + 3a
j = 30 + 3 × 45
j = 30 + 135
j = 165
Aiden's weight is 45 pounds, Jonathan's weight is 165 pounds
Answer:
81.86%
Step-by-step explanation:
We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.
First of all we will find z-score using z-score formula.
Now let us find z-score for 86.
Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.
Using normal distribution table we will get,
We will use formula to find the probability to find that a normal variable lies between two values.
Upon substituting our given values in above formula we will get,
Upon converting 0.81859 to percentage we will get
Therefore, 81.86% of final exam score will be between 68 and 86.