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Daniel [21]
3 years ago
14

Using the porpotions: WHIP= (Walks+Hits) divide by Innings Pitched. The pitcher wants to lower his WHIP from 1.53 to 1.3. How ma

ny more innings would he have to pitch without any walks or hits? Is this a reasonable goal? Explain your reasoning.
Mathematics
1 answer:
BaLLatris [955]3 years ago
7 0
This relates to ur previous question....there were 85 innings last question and 130 walks + hits which gave a whip score of 1.53......so if he wants to lower his whip (without changing his walks or hits) to 1.3....
130 / (85 + x) = 1.3....multiply both sides by (85 + x)
130 = 1.3(85 + x)
130 = 110.50 + 1.3x
130 - 110.50 = 1.3x
19.5 = 1.3x
19.5 / 1.3 = x
15 = x

he would have to pitch 15 more innings without any walks or hits to reach a goal of 1.3 whip score. This is not a reasonable goal.....have u ever seen a baseball game where a pitcher pitched 15 innings without any walks or hits ? And since a game consists of 9 innings....he would be 3 innings short of 2 games (thats if they dont go into overtime), and he would have to pitch a no hitter....and that is highly unlikely.
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$152.28

Step-by-step explanation:

In order to determine the markup, multiply the original price by 62%

94 x 62% = 94 x .62 = 58.28

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3 years ago
Honors Geometry QUiz- Please Help
viktelen [127]

Answer:

QUESTION 2:

The area of the polygon is 21.5 units²

QUESTION 3:

The area of the combined figure is 12 + 18.25·π units²

QUESTION 4:

The area of the figure is 40 units²

QUESTION 5

The area of the compost shape is 40 in.²

Step-by-step explanation:

QUESTION 2.

The area of the polygon is given as follows;

The area of two triangles + The area of trapezoid

1/2×6×3 + 1/2×3×4 + 1/2×(6 + 7) = 21.5 units²

QUESTION 3

The diameter of the semicircle, 'd', is given as follows;

d = √(8² + 3²) = √73

The area of the semicircle = π·d²/4 = π·73/4 = 18.25·π unit²

The area of the right triangle beside the semicircle is given as follows;

1/2 × 8 × 3 = 12 unit²

The area of the combined figure = The area of the semicircle + The area of the right triangle

∴ The area of the combined figure = 12 + 18.25·π units²

QUESTION 4:

The dimensions of the parallelogram are;

Width, W = √(4² + 1²) = √(17)

Length, L = √(8² + 2²) = √(68)

The area of the parallelogram, A_R = W × L = √(17) × √(68) = √(17 × 68) = 34

The area of the parallelogram, A_R = 34 units²

The length of a parallel side, 'A', of the parallelogram is A = 6

The height of the parallelogram, is h = 1

The area of the parallelogram is A = B·h = 6 × 1 = 6 units²

The area of the figure is the sum of the area of the rectangle and the area of the parallelogram

Therefore, we have;

The area of the figure = 34 units² + 6 units² = 40 units²

QUESTION 5

The composite shape consists of a triangle and a rectangle

Therefore, we have;

The area of the rectangle = 5 in. × 7 in. = 35 in.²

The area of the triangle = 1/2 × 2 in. × 5 in. = 5 in.²

The area of the compost shape = The area of the rectangle + The area of the triangle

∴ The area of the compost shape = 35 in.² + 5 in.² = 40 in.²

The area of the compost shape = 40 in.².

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Answer:

the correct answer is B

Step-by-step explanation:

hope this helps

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2 years ago
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