Humood's house is (-5,7)
1 answer:
The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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Answer:
1, 2, 6
Step-by-step explanation:
The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:
Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts
1) For x < 90:
From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%
Option 1 is correct
2) For x > 130:
From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%
Option 2 is correct
3) For x > 190:
From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%
Option 3 is incorrect
4) For x < 130:
For x > 100:
From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%
Option 4 is incorrect
5) For x = 130:
Option 5 is incorrect
6) For x = 130:
Since 1.5 is between 1 and 2, option 6 is correct
D. 4 19/24
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