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Inessa05 [86]
2 years ago
10

What is the GCF of 12a^4b^2 - 3a^2b^5 ??

Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
8 0
<h2>GCF for the numerical part is 3 .</h2>

hope it helps~

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Find the length of line segment x.<br> A)<br> 4<br> B)<br> 5<br> 6<br> 7
Korolek [52]

Answer:

x = 6

Step-by-step explanation:

4(x + 4) = 5(3 + 5) => Intersecting Secants Theorem

4(x + 4) = 5(8)

Open the bracket

4x + 16 = 40

Subtract 16 from each side

4x + 16 - 16 = 40 - 16

4x = 24

4x/4 = 24/4

x = 6

4 0
3 years ago
Malika bought 7 new pairs of shoes. Each pair of shoes cost the same amount and was discounted by $21.75.
ikadub [295]

Multiply the amount of the discount by pairs of shoes:

21.75 x 7 = 152.25

Add that to the total she spent:

152.25 + 654.50 = 806.75

Divide the total by pairs of shoes:

806.75/7 = $115.25

The answer is C. $115.25

3 0
2 years ago
Given f (x) = 6 (1- x), what is the value of f (-5)?
natta225 [31]

Answer: 36

Step-by-step explanation:

1) substitute x for 6

get f(-5)=6(1--5)= 6+30=46

5 0
3 years ago
Read 2 more answers
Find three consecutive interferes whose sum is 96
Ronch [10]

Answer:

31, 32, 33

Step-by-step explanation:

x + x + 1 +x + 2 = 96

3x + 3 = 96

      -3    -3

3x = 93

/3     /3

x = 31

31 is the first number, so since they’re consecutive, the second two numbers are 32 and 33.

6 0
3 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
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