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3 years ago

Hey can you please help me posted picture of question

Mathematics

2 answers:

mixas84 [53]3 years ago

8 0

D) always less than. hope that help..

mr_godi [17]3 years ago

3 0

It could be B but double check to be sure

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David currently has $270 in his bank account and

Ivan

Answer:

12 gifts

Step-by-step explanation:

set the equation to 270-15x=330-20x

6 0

3 years ago

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Please Help!! 100 POINTS!!!

pychu [463]

H(x) = 16 - x
plug -4 in as x

H(x) = 16 - (-4)
solve 16 - (-4)

H(x) = 20

your answer is A)20

3 0

2 years ago

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Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

3 years ago

Hudson bought several cans of tennis balls. Each contained both green and yellow tennis balls. He purchased 10 green tennis ball

svp [43]

Answer:

Step-by-step explanation:

We don't know how many cans Hudson bought.

We are told, he bought:

10 green balls

5 yellow balls

Each can has 2 green balls, so there are:

10/2 = 5 cans

How can we distribute 5 yellow balls in 5 cans? Simple!

5/5 = 1

Put 1 yellow ball in each can.

So, each can will have:

2 green + 1 yellow (balls)

So, each can has 1 yellow ball

6 0

2 years ago

What do you get when you add 1999+1999 together?

pentagon [3]

It is 3,998 because if u add it together it is like adding 2,000 +2,000-2

4 0

3 years ago

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