What is 27/125 equal to

This probability distribution shows the typical distribution of pitches thrown to a batter in a given at bat in a baseball game.

Vikki [24]

Answer:

0.35

Step-by-step explanation:

This probability distribution is shown below:

Pitch 1 2 3 4 5

Frequency 15 20 40 15 10

Probability 0.15 0.2 0.4 0.15 0.1

The probability that the pitcher will throw fewer than 3 pitches to a batter = P(X < 3)

X is the number of pitches thrown. Therefore:

P(X < 3) = P(X = 1) or P(X = 2)

The additive rule pf probability states that if two events X and Y are dependent events, the probability of X or Y occurring is the sum of their individual probability.

P(X < 3) = P(X = 1) or P(X = 2) = P(X = 1) + P(X = 2) = 0.15 + 0.2 = 0.35

The probability that the pitcher will throw fewer than 3 pitches to a batter = 0.35

8 0

2 years ago

Ten times the sum of half a number x and 8 is 12.

Diano4ka-milaya [45]

BREAKDOWN

10 times: 10 ×

The sum of: ( ) + ( )

which would be : 10 × ( ) + ( )

half a number: 10 × ( X/2 ) + ( )

and 8: 10 × ( X/2 ) + ( 8 )

is 12 : 10 × ( X/2 + 8 ) = 12

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SIMPLIFICATION

Multiply ten: 10 X (X/2 + 8) = 12

which equals to: 5x + 80 = 12

bring the like terms together: 5x = 80 - 12

Equals: 5x = 78

Find out X:. x = 78/5

FINAL ANSWER:. X = 15.9

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6 0

3 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

2 years ago

How do I simplify this

velikii [3]

512x^3 would be the answer

7 0

3 years ago

Food costs are expected to rise 6% each month for the next year. Which series correctly depicts the cost (to the nearest cent) f

alexira [117]

For this case we have the following equation:
y = 150 * (1.06) ^ t
For the first month we have:
y = 150 * (1.06) ^ 1
y = 159 $
For the second month we have:
y = 150 * (1.06) ^ 2
y = 168.54 $
For the third month we have:
y = 150 * (1.06) ^ 1
y = 178.65 $
Answer:
d. $ 159.00 + $ 168.54 + $ 178.65

4 0

2 years ago

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