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Maurinko [17]
3 years ago
11

9x² - 12x + 4 pls need ans.​

Mathematics
1 answer:
kolezko [41]3 years ago
4 0

Step-by-step explanation:

\tt{9 {x}^{2}  - 12x + 4}

Here, the second order of polynomial ax² + bx + c is factorized and expressed as the product of two linear factors. First, we have to find the two numbers that adds to 12 and if we multiply those numbers , we get 36 ( The two numbers should be 6 and 6 ).

⟶ \tt{9 {x}^{2}  - (6 + 6)x + 4}

⟶ \tt{9 {x}^{2}  - 6x - 6x + 4} { Distribute x through the parentheses )

⟶ \tt{ \underbrace{9 {x}^{2}  - 6x}} \:  -  \underbrace{6x + 4}

⟶ \tt{3x(3x - 2) - 2(3x - 2)}

⟶ \tt{(3x - 2)(3x - 2)}

⟶ \tt{ {(3x - 2)}^{2} }

\pink{ \boxed{ \boxed{ \tt{ Our \: final \: answer :  \boxed{ \underline{ \tt{{(3x - 2)}^{2} }}}}}}}

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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) All human blood can be typed as one of O, A, B, or AB. The distribution of the type varies a bit with race. For African-Americ
ivann1987 [24]

Answer:

The correct option is 1 - [(0.8)¹⁰+10*0.2*(0.8)⁹]= 0.6242

Step-by-step explanation:

Hello!

Given the distribution of probabilities for blood types for African-Americans:

O: 0.4

A: 0.2

B: 0.32

AB: 0.08

A random sample of 10 African-American is chosen, what is the probability that 2 or more of them have Type A blood?

Let X represent "Number of African-Americans with Type A blood in a sample of 10.

Then you have two possible outcomes,

"Success" the person selected has Type A blood, with an associated probability p= 0.2

"Failure" the selected person doesn't have Type A blood, with an associated probability q= 0.8

(You can calculate it as "1-p" or adding all associated probabilities of the remaining blood types: 0.4+0.32+0.08)

Considering, that there is a fixed number of trials n=10, with only two possible outcomes: success and failure. Each experimental unit is independent of the rest and the probability of success remains constant p=0.2, you can say that this variable has a Binomial distribution:

X~Bi(n;p)

You can symbolize the asked probability as:

P(X≥2)

This expression includes the probabilities: X=2, X=3, X=4, X=5, X=6, X=7, X=8, X=9, X=10

And it's equal to

1 - P(X<2)

Where only the probabilities of X=0 and X=1 are included.

There are two ways of calculating this probability:

1) Using the formula:

P(X)= \frac{n!}{(n-X)!X!} *p^{x} * q^{n-x}

With this formula, you can calculate the point probability for each value of X=x₀ ∀ x₀=1, 2, 3, 4, 5, 6, 7, 8, 9, 10

So to reach the asked probability you can:

a) Calculate all probabilities included in the expression and add them:

P(X≥2)= P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + X=10

b) Use the complement rule and calculate only two probabilities:

1 - P(X<2)= 1 - [P(X=0)+P(X=1)]

2) Using the tables of the binomial distribution.

These tables have the cumulative probabilities listed for n: P(X≤x₀)

Using the number of trials, the probability of success, and the expected value of X you can directly attain the corresponding cumulative probability without making any calculations.

>Since you are allowed to use the complement rule I'll show you how to calculate the probability using the formula:

P(X≥2) = 1 - P(X<2)= 1 - [P(X=0)+P(X=1)] ⇒

P(X=0)= \frac{10!}{(10-)0!0!} *0.2^{0} * 0.8^{10-0}= 0.1074

P(X=1)= \frac{10!}{(10-1)!1!} *0.2^{1} * 0.8^{10-1}= 0.2684

⇒ 1 - (0.1074+0.2684)= 0.6242

*-*

Using the table:

P(X≥2) = 1 - P(X<2)= 1 - P(X≤1)

You look in the corresponding table of n=10 p=0.2 for P(X≤1)= 0.3758

1 - P(X≤1)= 1 - 0.3758= 0.6242

*-*

Full text in attachment.

I hope it helps!

8 0
3 years ago
Newborn babies: A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights
Fed [463]

Answer:

a) 615

b) 715

c) 344

Step-by-step explanation:

According to the Question,

  • Given that,  A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 732 babies born in New York. The mean weight was 3311 grams with a standard deviation of 860 grams

  • Since the distribution is approximately bell-shaped, we can use the normal distribution and calculate the Z scores for each scenario.

Z = (x - mean)/standard deviation

Now,

For x = 4171,  Z = (4171 - 3311)/860 = 1  

  • P(Z < 1) using Z table for areas for the standard normal distribution, you will get 0.8413.

Next, multiply that by the sample size of 732.

  • Therefore 732(0.8413) = 615.8316, so approximately 615 will weigh less than 4171  

 

  • For part b, use the same method except x is now 1591.    

Z = (1581 - 3311)/860 = -2    

  • P(Z > -2) , using the Z table is 1 - 0.0228 = 0.9772 . Now 732(0.9772) = 715.3104, so approximately 715 will weigh more than 1591.

 

  • For part c, we now need to get two Z scores, one for 3311 and another for 5031.

Z1 = (3311 - 3311)/860 = 0

Z2 = (5031 - 3311)/860= 2  

P(0 ≤ Z ≤ 2) = 0.9772 - 0.5000 = 0.4772

  approximately 47% fall between 0 and 1 standard deviation, so take 0.47 times 732 ⇒ 732×0.47 = 344.

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