Answer: 0.9332.
Step-by-step explanation:
Claim : College Algebra final exam score of engineering majors equal to 88.
Given that : The test statistic is z equals to 1.50.
To find the p-value (Probability value), we use standard normal distribution table, and search the p-value corresponds to the z-score.
In a Standard Normal Distribution Table below, the p-value corresponds z equals 1.5 is 0.9332.
Hence, the p-value is 0.9332.
Answer: Anything between 0 and 10, excluding both endpoints.
In terms of symbols we can say 0 < w < 10 where w is the width.
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Explanation:
You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.
w = width
2w = twice the width
2w-5 = five less than twice the width = length
So,
which lead to
area = length*width
area = (2w-5)*w
area = 2w^2-5w
area < 150
2w^2 - 5w < 150
2w^2 - 5w - 150 < 0
To solve this inequality, we will solve the equation 2w^2-5w-150 = 0
Use the quadratic formula. Plug in a = 2, b = -5, c = -150
Ignore the negative solution as it makes no sense to have a negative width.
The only practical root is w = 10.
If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.
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Based on that root, we need to try a sample value that is to the left of it.
Let's say we try w = 5.
2w^2 - 5w < 150
2*5^2 - 5*5 < 150
25 < 150 ... which is true
This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.
Now try something to the right of 10. I'll pick w = 15
2w^2 - 5w < 150
2*15^2 - 5*15 < 150
375 < 150 ... which is false
It means w > 10 leads to 2w^2-5w < 150 being false.
Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.
Answer:
875 ft²
Step-by-step explanation:
Finding area of similar rectangles:
Scale factor = EF : AB
= 25 : 5
= 5 : 1
Area or rectangle EFGH = 875 ft²
Answer:
To find the scale factor of the enlargement, compare the distance between a pair of corresponding points from both shapes.
<u>Shape K</u>
A = (4, 7)
B = (7, 7)
C = (7, 4)
D = (5, 5)
Horizontal distance between A (4, 7) and B (7, 7) = 3 units
<u>Shape L</u>
A' = (0, 11)
B' = (9, 11)
C' = (9, 2)
D' = (3, 5)
Horizontal distance between A' (0, 11) and B' (9, 11) = 9 units
9 ÷ 3 = 3
Therefore, Shape L is an enlargement of Shape K by scale factor 3.
To find the center of dilation (enlargement), draw two lines through 2 corresponding points (e.g. A and A', B and B') - the point of intersection of these lines is the center of dilation.
Therefore, the center of enlargement is (6, 5) (refer to the second attached image).
