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2 years ago

Last season, when a football team was doing poorly, weekly attendance averaged 42,000. This season

Mathematics

1 answer:

fiasKO [112]2 years ago

4 0

My teacher always told me: use the percent change formula!!!

Here it is: new value-old value/ old value * 100%

This strategy will ALWAYS work for percent increase AND decrease: let's try it on this problem!

56700-42000/42000 * 100%

14,700/42000 = 100(x)

100* 14700 = 1470000

1470000/42000= 35

35/100 = 35%

Therfore, our percent increase is 35%!

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A study indicates that 62% of students have have a laptop. You randomly sample 8 students. Find the probability that between 4 a

Scrat [10]

Answer:

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a laptop, or they do not. The probability of a student having a laptop is independent from other students. So we use the binomial probability distribution to solve this question.

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.

A study indicates that 62% of students have have a laptop.

This means that $n = 0.62$

You randomly sample 8 students.

This means that $n = 8$

Find the probability that between 4 and 6 (including endpoints) have a laptop.

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157$

$P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815$

$P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297$

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269$

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

3 0

3 years ago

there are 3 teams of 18 employees working today company policy says that we need to have one supervisor for every 8 employees on

Vadim26 [7]

The <span> answer is 6.75. Since if you divide the number by 8 you get 6.75.</span>

8 0

2 years ago

Read 2 more answers

g Bonus: Assume that among the general pediatric population, 7 children out of every 1000 have DIPG (Diffuse Intrinsic Pontine G

patriot [66]

Answer:

0.9586

Step-by-step explanation:

From the information given:

7 children out of every 1000 children suffer from DIPG

A screening test designed contains 98% sensitivity & 84% specificity.

Now, from above:

The probability that the children have DIPG is:

$\mathbf{P(positive) = P(positive \ | \ DIPG) \times P(DIPG) + P(positive \ | \ not \DIPG)\times P(not \ DIPG)}$ $= 0.98\imes( \dfrac{7}{1000}) + (1-0.84) \times (1 - \dfrac{7}{1000})$