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nikitadnepr [17]
3 years ago
7

Jose and Franco were playing a video game if Jose score 782,076 points and Franco scored 811,968 point what is the difference be

tween the points scored
Mathematics
1 answer:
posledela3 years ago
7 0

Answer:

29,892

Step-by-step explanation:

811,968-782,076

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the simple intrest earned on $500,000 invested for eighteen months is $90,000 what was the interest rate?​
kvv77 [185]

Answer:

am comin to you I going to work it

7 0
3 years ago
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Solve for y:<br>5(x + y) = 20 + 3x​
son4ous [18]
5(x+y)= 20+ 3x
5x+5y=20+3x
5x+5y-3x=20+3x-3x
2x+5y=20
2x+5y-2x=20-2x
5y=-2x+20
5y/5=-2x/5+20/5
y=-2/5x+4

ANSWER: y=-2/5x+4
5 0
3 years ago
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Find the average rate of change of the function between the given values of x. y = 9+ 5x + 0.5x2 between x = 2 and x = 4.
svet-max [94.6K]

Answer:

The average rate of change is 8.

Step-by-step explanation:

The formula to calculate the average rate of change of a function F(x) is:

\frac{F(b)-F(a)}{b-a}

In this case, F(x) = 0.5x^{2} +5x+9

a=2 and b=4

You have to evaluate x=2 (which is a in the formula) and x=4 (which is b in the formula) in the function.

In order to obtain F(b) and F(a) you have to replace x=4 and x=2 in the given function:

F(b) = (0.5)4^{2} + 5(4) +9= 37

F(a) = (0.5)2^{2} + 5(2)+9=21

\frac{F(b)-F(a)}{b-a} = \frac{37-21}{4-2}=\frac{16}{2} = 8

The answer is 8.

3 0
3 years ago
a local charity held a crafts fair selling donated, handmade items. Total proceeds from the sale were 1875. a total of 95 items
Anastasy [175]
55 items sold for $15 each.
40 sold for $25 each.
3 0
3 years ago
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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
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