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Strike441 [17]
3 years ago
14

Apply the distributive property to factor out the greatest common factor. 56 + 32 =

Mathematics
1 answer:
valentina_108 [34]3 years ago
4 0

Answer:

8 * (7 + 4)

See process below

Step-by-step explanation:

We start by writing each number in PRIME factor form:

56 = 2 * 2 * 2 * 7

32 = 2 * 2 * 2 * 2 * 2

Notice that the factors that are common to BOTH numbers are 2 * 2 * 2 (the product of three factors of 2).Therefore we see that the greatest common factor for the given numbers is : 2 * 2 * 2 = 8

Using this, we can write the two numbers as the product of this common factor (8) times the factors that are left on each:

56 = 8 * 7

32 = 8 * 2 * 2 = 8 * 4

We can then use distributive property to "extract" that common factor (8) from the given addition as shown below:

56 + 32

8 * 7 + 8 * 4

8 * (7 + 4)

8 * (11)

88

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Y = (x)(x) is a parent function that makes a graph of a(n)
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Answer:

parabola

Step-by-step explanation:

y=x^{2} is the parent function of a parabola

4 0
3 years ago
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4. Assume that the chances of a basketball player hitting a 3-pointer shot is 0.4 and the probability of hitting a free-throw is
vovikov84 [41]

Answer:

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

Step-by-step explanation:

For each 3-pointer shot, there are only two possible outcomes. Either the player makes it, or the player does not. The same is valid for free throws. This means that both the number of 3-pointers and free throws made are given by binomial distributions.

Since 3-pointers and free throws are independent, first we find the probability of making exactly 3 3-pointers out of 10, then the probability of making exactly 5 free throws out of 10, and then the probability that the player will make exactly 3 3-pointers and 5-free throws is the multiplication of these probabilities.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of making 3 3-pointers out of 10:

The chances of a basketball player hitting a 3-pointer shot is 0.4, which means that p = 0.4. So this is P(X = 3) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.21499

Probability of making 5 free throws out of 10:

The probability of hitting a free-throw is 0.65, which means that p = 0.65. The probability is P(X = 5) when n = 10.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{10,5}.(0.65)^{5}.(0.35)^{5} = 0.15357

Calculate the probability that the player will make exactly 3 3-pointers and 5-free throws.

0.21499*0.15537 = 0.0334

0.0334 = 3.34% probability that the player will make exactly 3 3-pointers and 5-free throws.

5 0
3 years ago
Type the correct answer in each box. If necessary, use / for the fraction bar(s).
Tom [10]

Given:

The values are a=0.\overline{6},b=0.75.

To find:

The values of ab and \dfrac{a}{b}.

Solution:

We have,

a=0.\overline{6}

a=0.666...                     ...(i)

Multiply both sides by 10.

10a=6.666...                 ...(ii)

Subtracting (i) from (ii), we get

10a-a=6.666...-0.666...

9a=6

a=\dfrac{6}{9}

a=\dfrac{2}{3}

And,

b=0.75

b=\dfrac{75}{100}

b=\dfrac{3}{4}

Now, the product of a and b is:

ab=\dfrac{2}{3}\times \dfrac{3}{4}

ab=\dfrac{1}{2}

The quotient of a and b is:

\dfrac{a}{b}=\dfrac{\dfrac{2}{3}}{\dfrac{3}{4}}

\dfrac{a}{b}=\dfrac{2}{3}\times \dfrac{4}{3}

\dfrac{a}{b}=\dfrac{8}{9}

Therefore, the required values are ab=\dfrac{1}{2} and \dfrac{a}{b}=\dfrac{8}{9}.

6 0
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What’s the answer to this question
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Answer:

ab

Step-by-step explanation:

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Subtract 9 from 3:

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Add -6 to 7:

ab

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