Question Please help. Alina spent no more than $45 on gas for a road trip. The first gas station she used charged $3.50 per gall
1 answer:
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Answer:
Shift 2 unit left
Flip the graph about y-axis
Stretch horizontally by factor 2
Shift vertically up by 2 units
Step-by-step explanation:
Given:
Parent function:
Transformation function:
Take -2 common from transform function f(x)
Now we see the step-by-step translation
Shift 2 unit left ( x → x+2 )
Flip the graph about y-axis ( (x+2) → - (x+2) )
Stretch horizontally by factor 2 [ -x(x+2) → -2(x+2) ]
Shift vertically up by 2 units [ f(x) → f(x) + 2 ]
Simplify the function:
Hence, Using four step of transformation to get new function
Answer:
(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.
(b) The fraction of the calls last more than 5.30 minutes is 0.1271.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.
(e) The time is 5.65 minutes.
Step-by-step explanation:
We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.
Let X = <u><em>the length of the calls, in minutes.</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time = 4.5 minutes
= standard deviation = 0.7 minutes
(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X 4.50 min)
P(X < 5.30 min) = P( <
) = P(Z < 1.14) = 0.8729
P(X 4.50 min) = P(
) = P(Z
0) = 0.50
The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = <u>0.3729</u>.
(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)
P(X > 5.30 min) = P( >
) = P(Z > 1.14) = 1 - P(Z
1.14)
= 1 - 0.8729 = <u>0.1271</u>
The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.
(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X 5.30 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 5.30 min) = P(
) = P(Z
1.14) = 0.8729
The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = <u>0.1109</u>.
(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X 4.00 min)
P(X < 6.00 min) = P( <
) = P(Z < 2.14) = 0.9838
P(X 4.00 min) = P(
) = P(Z
-0.71) = 1 - P(Z < 0.71)
= 1 - 0.7612 = 0.2388
The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.
Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = <u>0.745</u>.
(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;
P(X > x) = 0.05 {where x is the required time}
P( >
) = 0.05
P(Z > ) = 0.05
Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;
x = 4.5 + 1.15 = 5.65 minutes.
SO, the time is 5.65 minutes.