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Gemiola [76]
2 years ago
12

3(x-5)² +14(x-5)-24

Mathematics
2 answers:
Pani-rosa [81]2 years ago
8 0

Answer:

3x^{2} -16x-19

Step-by-step explanation:

If we are going to be simplying this problem, then we need to take it step by step.

Start off by using the distributive property, and then combining like terms. Keep in mind the square, and make sure that gets sorted out correctly.

azamat2 years ago
3 0

Answer:

(3x - 19) • (x + 1)

Step-by-step explanation:

STEP

1

:

Equation at the end of step 1

((3•((x-5)2))+14•(x-5))-24

STEP

2

:

Equation at the end of step 2

(3 • (x - 5)2 + 14 • (x - 5)) - 24

STEP

3

:

Trying to factor by splitting the middle term

Factoring 3x2-16x-19

The first term is, 3x2 its coefficient is 3 .

The middle term is, -16x its coefficient is -16 .

The last term, "the constant", is -19

Step-1 : Multiply the coefficient of the first term by the constant 3 • -19 = -57

Step-2 : Find two factors of -57 whose sum equals the coefficient of the middle term, which is -16 .

-57 + 1 = -56

-19 + 3 = -16 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -19 and 3

3x2 - 19x + 3x - 19

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (3x-19)

Add up the last 2 terms, pulling out common factors :

1 • (3x-19)

Step-5 : Add up the four terms of step 4 :

(x+1) • (3x-19)

Which is the desired factorization

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If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ is approximately a normal distribution
FrozenT [24]

Answer:

Yes we can conclude.

Step-by-step explanation:

The sampling distribution of \hat{p} can be approximated as a Normal Distribution only if:

np and nq are both equal to or greater than 10. i.e.

  • np ≥ 10
  • nq ≥ 10

Both of these conditions must be met in order to approximate the sampling distribution of \hat{p} as Normal Distribution.

From the given data:

n = 50

p = 0.80

q = 1 - p = 1 - 0.80 = 0.20

np = 50(0.80) = 40

nq = 50(0.20) = 10

This  means the conditions that np and nq must be equal to or greater than 10 is being satisfied. So, we can conclude that the sampling distribution of pˆ is approximately a normal distribution

6 0
3 years ago
In a school of 330 students, 85 of them are in a drama club, 200 of them are in a sports team and 60 of them do both
Neko [114]

Answer:

105/330 = 21/66 = 7/22

7/22 is the answer I think

Step-by-step explanation:

n(U) = 330

n(A) = 85

n(B) = 200

n(A n B) = 60

Now,

n(A U B) = n(A) + n(B) - n(A n B)

or, n(A U B) = 85 + 200 - 60

or, n(A U B) = 285 - 60 = 225

Now,

n(A U B) compliment = n(U) - n(A U B)

n(A U B) compliment = 330 - 225

so, n(A U B) compliment = 105

Now,

probability = 105/330

8 0
2 years ago
dear brainly, I hope you see this, Please be aware of when you go and delete a persons account and or questions and answers. Be
Scrat [10]

Answer:

So true bro

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Solve the inequality for x. Show each step of the solution.<br> 12X&gt;9(2X-3)-15
agasfer [191]
12x > 18x - 27 - 15
12x > 18x - 42
12x - 18x > -42
-6x > -42
x < -42/-6
two negatives make a positive so now its x < 42/6
Lastly divide 42/6 so it can equal 7

Answer: X < 7.
6 0
3 years ago
Read 2 more answers
The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab
LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.  

From the central limit theorem we have the formula:

z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }

a) z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

\therefore P(\bar X\:>\:1224)=0.0082

b) z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00

The area to the left of z=3.00 is 0.9987

\therefore P(\bar X\:

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

P(1200\:

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

5 0
3 years ago
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