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3 years ago

The product of 84.12 and which number will have three decimal places?

Mathematics

2 answers:

Irina18 [472]3 years ago

6 0

Answer:

19.3

Step-by-step explanation:

Setler79 [48]3 years ago

6 0

Answer:

9.13

Step-by-step explanation:

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Find a function that describes the sequence. Use y for the term in the sequence and n for its place in the list. Then use the fu

sammy [17]

B. y=n+20; 25,26,27 because if you plug in 1 into the equation it equals 21. 2+20=22, etc. so to find the next three terms of the equation plug in 5: y=5+20=25; plug in 6: y=6+20=26; and plug in 7: y=7+20=27

6 0

3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

Lubov Fominskaja [6]

Answer:

The required sample size for the new study is 801.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What is the required sample size for the new study?

This is n for which M = 0.03. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.25*0.75)}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{1.96\sqrt{0.25*0.75}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.25*0.75}}{0.03})^2$

$n = 800.3$

Rounding up:

The required sample size for the new study is 801.

4 0

3 years ago

Savannah had 400 chocolate chips. She saved x chips for a garnish, and mixed the rest of the chocolate chips evenly into 3 pans

scoundrel [369]

Answer:

100

Step-by-step explanation:

400 divide by 4 =100

6 0

4 years ago

Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used 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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

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

In which

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

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

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

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

7 0

4 years ago

PLEASE HELP ME WITH THIS QUESTION

Karo-lina-s [1.5K]

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the fastest driver is Evans

5 0

4 years ago