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Umnica [9.8K]
3 years ago
9

Twenty percent of the 640 pitches Kendall threw last year were strikes. Which table can be used to find the number of strikes sh

e threw?Number of PitchesA 3-column table with 6 rows. Column 1 has entries 20 percent, 20 percent, 20 percent, 20 percent, 20 percent = 100 percent. Column 2 has entries 0.2, 0.2, 0.2, 0.2, 0.2 = 1. Column 3 has entries 128, 128, 128, 128, 128 =640.Number of PitchesA 3-column table with 6 rows. Column 1 has entries 20 percent, 20 percent, 20 percent, 20 percent, 20 percent = 100 percent. Column 2 has entries 0.2, 0.2, 0.2, 0.2, 0.2 = 1. Column 3 has entries 256, 256, 256, 256, 256 = 1,280.Number of PitchesA 3-column table with 5 rows. Column 1 has entries 25 percent, 25 percent, 25 percent, 25 percent = 100 percent. Column 2 has entries 0.25, 0.25, 0.25, 0.25 = 1. Column 2 has entries 160, 160, 160, 160 = 640.Number of PitchesA 3-column table with 5 rows. Column 1 has entries 25 percent, 25 percent, 25 percent, 25 percent = 100 percent. Column 2 has entries 0.25, 0.25, 0.25, 0.25 = 1. Column 2 has entries 320, 320, 320, 320 = 1,280.
Mathematics
1 answer:
fenix001 [56]3 years ago
5 0

Answer: A 12 Percent

Step-by-step explanation:

good luck?

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a triangle pyramid with an equilateral base has a side length of 10 centimeters and a surface area of 214.5 square meters. find
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Area of the base = 1/2 * 10 *  5sqrt3 =  25 sqrt3

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A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
bonufazy [111]

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

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