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natali 33 [55]
3 years ago
6

Does anybody know 5.4.7: Teenagers CodeHs. Please help I am stuck in this

Mathematics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

The program in Python is as follows:

age = int(input("Age: "))

if age>=13 and age <=19:

   print("Teenager")

else:

   print("Not a teenager")

Step-by-step explanation:

This gets input for age

age = int(input("Age: "))

This checks if age is between 13 and 19 (inclusive)

if age>=13 and age <=19:

If yes, the age is teenage

   print("Teenager")

If otherwise, the age is not teenage

else:

   print("Not a teenager")

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Find all possible values of b such that 4x² + bx + 3 can be factored. Check all that apply.
aleksklad [387]

The possible values of b are 7 and 8

<h3>How to determine the possible values of x?</h3>

The expression is given as:

4x² + bx + 3

Next, we test the options to determine the values of b

<u>Option 1: b = 13</u>

So, we have:

4x² + 13x + 3 ---- this cannot be factorized

<u>Option 2: b = 7</u>

So, we have:

4x² + 7x + 3  

Expand

4x² + 4x + 3x + 3

Factorize

4x(x +1) + 3(x + 1)

Factor out x + 1

(4x + 3)(x + 1) ------ this can be factorized

<u>Option 3: b = 8</u>

So, we have:

4x² + 8x + 3

Expand

4x² + 6x + 2x + 3

Factorize

2x(2x +3) + 1(2x + 3)

Factor out 2x + 3

(2x + 3)(2x + 1) ------ this can be factorized

<u>Option 4: b = 1</u>

So, we have:

4x² + x + 3 ---- this cannot be factorized

Hence, the possible values of b are 7 and 8

Read more about factorized expressions at:

brainly.com/question/723406

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3 0
2 years ago
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls.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We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls.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

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Step-by-step explanation:

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