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2 years ago

Four students are competing in a race. In how many different combinations could the four students place in the race?

1 answer:

4 0

"Combinations" isn't really the right word, because a combination doesn't take order of elements into account, and the outcome of a race certainly does depend on order. "Permutations" would be the correct term.

If 4 students are competing, then the race has

4! = 4 • 3 • 2 • 1 = 24

possible outcomes.

You might be interested in

Which represents the solution(s) of the equation x^2= 289

PilotLPTM [1.2K]

Answer:

x² is equal to 289 so x is obviously the square root of 289 which is 17

5 0

2 years ago

Read 2 more answers

Classify the system.

belka [17]

Consistent and independent

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3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 335335 babies were born

strojnjashka [21]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

Step-by-step explanation:

Confidence Interval for the proportion:

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 335, \pi = \frac{268}{335} = 0.8$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 - 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.7437$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8 + 2.575\sqrt{\frac{0.8*0.2}{335}} = 0.8563$

For the percentage:

Multiplying the proportions by 100.

The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).

Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.

7 0

3 years ago

Two factory plants are making TV panels. Yesterday, Plant A produced 6000 panels. Seven percent of the panels from Plant A and 2

azamat

Answer:

1% of 4000 + 4% of B = 2% of (4000 + B)

40 + 0.04*B = 0.02( 4000 + B)

40 + 0.04* B = 80 + 0.02*B

40 + 0.04*B - 80 - 0.02B = 0

0.02*B - 40 = 0

0.02*B = 40

2B = 4000

B = 2000

Step-by-step explanation:

7 0

2 years ago

A pen stand costs ₹120. How many pen stand can be bought for ₹1440

siniylev [52]

1440 divided by 120 = 12 pen stands

hope it helps:))

6 0

3 years ago