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elena55 [62]
3 years ago
6

A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7. Your friend concludes that red cards and even numbers must be mutually exclusive.Is your friend correct? Explain.
Mathematics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that a bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag.

This is done four times. Two red cards are drawn, numbered 31 and 19, and two blue cards are drawn, numbered 22 and 7.

From the above we cannot conclude that red cards and even numbers are mutually exclusive

Just drawing two red cards and because the two happen to be odd we cannot generalize the red cards have odd numbers.

This might have occurred due to simple chance from a comparatively large number of 40 cards.

Suppose say we have red cards 20, and 19 red 1 blue.

Then drawing 2 from 19 red cards have more probability and this can occur by chance.

So friend's conclusion is wrong.

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The newly elected president needs to decide the remaining 7 spots available in the cabinet he/she is appointing. If there are 13
Alexxx [7]

Answer: 8648640 ways

Step-by-step explanation:

Number of positions = 7

Number of eligible candidates = 13

This can be done by solving the question using the combination Formula for selection in which we use the combination formula to choose 7 candidates amomg the possible 13.

The combination Formula is denoted as:

nCr = n! / (n-r)! * r!

Where n = total number of possible options.

r = number of options to be selected.

Hence, selecting 7 candidates from 13 becomes:

13C7 = 13! / (13-7)! * 7!

13C7 = 1716.

Considering the order they can come in, they can come in 7! Orders. We multiply this order by the earlier answer we calculated. This give: 1716 * 7! = 8648640

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3 years ago
What is the Answer to -14.5 + (-1.24)
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How do you change 20.25 into a fraction
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A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling
lara31 [8.8K]

Answer:

a) weight  of the car = 2816,1 lbs

b) 2773 lbs

Step-by-step explanation:

The equilibrium force is 490 lbs. That force keep the car at rest, then

∑ Fy  =  0     and  ∑Fx  =  0

Forces acting on the car:

The external force   490 lbs

weight of the car   uknown

Normal force

sin∠10°  =  0,174

cos∠10° = 0.985

∑Fx  =  0         mg*sin10°- 490 = 0      ∑Fy =  0      mg*cos10° - N  =  0

mg*0,174= 490

mg  =  490 / 0,174

mg = 2816,1 lbs

weight  of the car = 2816,1 lbs

The Normal force

mg*cos10° - N  =  0        2816,1 * 0,985 = N

N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway

3 0
3 years ago
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