Jon and Ade share some socks in the ratio 2:7 Jon gets 18 sock how many does Ade get

As part of a science experiment, Giovanni measured the distance a caterpillar traveled over different time intervals. In the fir

777dan777 [17]

Step-by-step explanation:

its b

8 0

4 years ago

21 less than tree-fifths of a number

Oxana [17]

(3/5)x - 21

hope this helps :)

6 0

2 years ago

An admiral, captain, and commander, all different, are to be chosen from a group of 10 Starfleet officers. How many different ch

zepelin [54]

Answer:

$10C1 *9C1*8C1 = 10*9*8=720$

Step-by-step explanation:

For this case in order to select the one admiral, captain and commander, all different. We are assuming that the order in the selection no matter, so we can begin selecting an admiral then a captain and then a commander.

So we have 10C1 ways to select one admiral since we want just one

Now we have remaining 9 people and we have 9C1 ways to select a captain since we want a captain different from the admiral selected first

Now we have remaining 8 people and we have 8C1 ways to select a commander since we want a commander different from the captain selected secondly.

The term nCx (combinatory) is defined as:

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

And by properties $nC1= n$

So then the number of possible way are:

$10C1 *9C1*8C1 = 10*9*8=720$

If we select first the captain then the commander and finally the admiral we have tha same way of select $10*9*8=720$

For all the possible selection orders always we will see that we have 720 to select.

5 0

3 years ago

Read 2 more answers

An urn contains three red balls and four blue balls. Draw two balls at random from the urn, without replacement. Compute the exp

mezya [45]

Step-by-step explanation:

in total we have 3+4 = 7 balls.

when we draw the first ball, the probability to draw a red ball is 3/7, and a blue ball 4/7.

when we draw the second ball, we have now only 6 balls in total.

the probabilty to draw a red back now depends also on the result of the first draw.

if the first ball was already red, then we have only a chance now of 2 out of 6.

if the first ball was blue, then we have now a chance of 3 out of 6.

so, the probably to draw at least 1 red ball in 2 draws is the probability of drawing one on the first draw plus the probability of drawing one on the second :

1 - probability to see 2 blue balls

1 - 4/7 × 3/6 = 1 - 12/42 = 30/42 = 0.714285714...

the expected number of red balls in 2 draws is

1 red in first red in first red in second

but not second and second but not first

1×(3/7 × 4/6) + 2×(3/7 × 2/6) + 1×(4/7 × 3/6) = 12/42 + 12/42 + 12/42 = 36/42 = 6/7 = 0.857142857 ≈ 0.8571

7 0

2 years ago

Now suppose that bigger cups are ordered and the machine’s mean amount dispensed is set at μ=12. Assuming we can precisely adjus

Elina [12.6K]

Answer:

σ should be adjusted at 0.5.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean 12.

Assuming we can precisely adjust σ, what should we set σtobe so that the actual amount dispensed is between 11 and 13 ounces, 95% of the time?

13 should be 2 standard deviations above the mean of 12, and 11 should be two standard deviations below the mean.

So 1 should be worth two standard deviations. Then

$2\sigma = 1$

$\sigma = \frac{1}{2}$

$\sigma = 0.5$

σ should be adjusted at 0.5.

4 0

2 years ago