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If 10 people apply for 3 jobs in how many ways can people be chosen for the jobs. 1. If the jobs are all the same. 2. If the jobs are all different. (please, please show work! Thank you!)

Answers

Answer 1

Answer:

Answer:

1. 120 ways

2. 720 ways

Step-by-step explanation:

When the order is important, we have a permutation.

When the order is not important, we have a combination.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_((n,x)) = (n!)/((n-x)!)$

Combinations formula:

$C_(n,x)$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

1. If the jobs are all the same.

Same jobs means that the order is not important. So

3 from a set of 10.

$C_(10,3) = (10!)/(3!(10-3)!) = 120$

120 ways

2. If the jobs are all different.

DIfferent jobs means that the order matters.

$P_((10,3)) = (10!)/((10-3)!) = 720$

720 ways

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Please answer these questions! Need help!

Answers

Answer:

Question 2 is $500 and Question 3 is 5

Step-by-step explanation:

Which expression shows the prime factorization of 40? 2 × 2 × 5 2 × 2 × 2 × 5 2 × 5 × 5 2 × 5 × 5 × 5

Answers

Answer:

2 × 2 × 2 × 5

Step-by-step explanation:

2 × 2 =4×2 =8×5=40

Answer:

2·2·2·5

Step-by-step explanation:

2·2=4·2=8·4=40

Please helppp it’s urgent , I’ll give a crown !!

Answers

Answer:

Slide to the left then rotate 90° to the right then slide down then rotate 180° to the left/ right.

Based on a Comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime-time TV live, instead of online, on DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch prime-time live.

Answers

Answer: Our required probability is 0.004672.

Step-by-step explanation:

Since we have given that

Number of adults = 7

Probability of getting adult will watch prime time TV live = 0.8

We need to find the probability that fewer than 3 of the selected adults watch prime time live.

We will use "Binomial Distribution":

here, n = 7

p = 0.8

So, P(X<3)=P(X=0)+P(X=1)+P(X=2)

So, it becomes,

$P(X=0)=(1-0.8)^7=0.2^7=0.0000128$

$P(X=1)=^7C_1(0.8)(0.2)^6=0.0003584\n\nP(X=2)=^7C_2(0.8)^2(0.2)^5=0.0043$

So, probability that fewer than 3 of the selected adult watch prime time live is given by

$0.0000128+0.0003584+0.0043=0.004672$

Hence, our required probability is 0.004672.

Final answer:

This problem relates to the binomial distribution and requires us to find the sum of binomial probabilities for 0, 1, and 2 successes (adults watching live TV) out of seven trials (the seven randomly selected adults).

Explanation:

This question is utilizing the concept of binomial distribution. The probability of a randomly selected adult watching prime-time TV live is 0.8. We want to find the probability that fewer than 3 out of 7 randomly selected adults watch prime-time live.

We find this by adding up the probabilities for 0, 1, and 2 adults watching live TV using the binomial distribution formula: P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k)), where C(n, k) denotes the number of combinations of n items taken k at a time, p is the probability of success, and n is the number of trials.

We get:

P(X=0) = C(7, 0) * (0.8^0) * ((1-0.8)^(7-0))
P(X=1) = C(7, 1) * (0.8^1) * ((1-0.8)^(7-1))
P(X=2) = C(7, 2) * (0.8^2) * ((1-0.8)^(7-2))

Adding these up will give the total probability that fewer than three adults out of seven watch prime-time live.

Learn more about Binomial Distribution here:

brainly.com/question/33656163

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