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2 years ago

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6(7a-10) how do I solve this

2 answers:

NeTakaya2 years ago

7 0

$6(7a-10) \\ \\ 6 \times 7a + 6 \times -10 \ / \ distribute \\ \\ 42a + 6 \times -10 \ / \ simplify \\ \\ 42a + -60 \ / \ simplify \\ \\ 42a - 60 \ / \ simplify \\ \\ Answer: \fbox {42a - 60}$

Masteriza [31]2 years ago

6 0

Solve the brackets, to do that you should multiply the numbers inside the bracket with the first number outside and in the right to the brackets.so,

6(7a-10)= 42a-60
hence the answer is, 42a-60

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2. It has been estimated that about 30% of frozen chickens are contaminated with enough salmonella bacteria to cause illness if

Darina [25.2K]

Using the binomial distribution, it is found that there is a 0.1111 = 11.11% probability that a certain crate has more than 4 contaminated chickens.

For each chicken, there are only two possible outcomes, either they are contaminated, or they are not. The chickens are independently selected, hence, the binomial distribution is used.

<h3>What is the binomial probability distribution formula?</h3>

The formula is:

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

About 30% of frozen chickens are contaminated, hence $p = 0.3$ .

Chickens are delivered to grocery stores in crates of 24, hence $n = 24$ .

The probability that a certain crate has <u>more than 4 contaminated</u> chickens is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then

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

$P(X = 0) = C_{24,0}.(0.3)^{0}.(0.7)^{24} = 0.0002$

$P(X = 1) = C_{24,1}.(0.3)^{1}.(0.7)^{23} = 0.0020$

$P(X = 2) = C_{24,2}.(0.3)^{2}.(0.7)^{22} = 0.0097$

$P(X = 3) = C_{24,3}.(0.3)^{3}.(0.7)^{21} = 0.0305$

$P(X = 4) = C_{24,4}.(0.3)^{4}.(0.7)^{20} = 0.0687$

Hence

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0020 + 0.0097 + 0.0305 + 0.0687 = 0.1111$

0.1111 = 11.11% probability that a certain crate has more than 4 contaminated chickens.

You can learn more about the binomial distribution at brainly.com/question/24863377

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