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sergejj [24]
3 years ago
9

Let A and B be subsets of a universal set U and suppose n(U) = 400, n(A) = 115, n(B) = 90, and n(A ∩ B) = 50. Find the number of

elements in the set.
Mathematics
1 answer:
babunello [35]3 years ago
8 0

By the inclusion/exclusion principle,

n(A\cup B)=n(A)+n(B)-n(A\cap B)=155

There are 400 elements in the universal set U, which means there are 400 - 155 = 245 elements not accounted for by A\cup B, or

n(U)=n(A\cup B)+n((A\cup B)^C)\implies n((A\cup B)^C)=245

That's all you can really determine from the given info. Considering the language of the problem, "Find the number of elements in the set", I find it hard to believe that the set it's talking about isn't mentioned.

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Step-by-step explanation:

the angle CDB is the supplementary angle to 9x+10 (angle ADC).

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so,

CDB + (4x + 10) + 50 = 180

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therefore,

(170 - 9x) + (4x + 10) + 50 = 180

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What is the measurement of angle ABC?
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A car's wheels spins at 1000 revolutions per minute. The diameter of the wheels is 23 inches. You want to know how fast the car
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72,220 inches per  minute

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3 0
3 years ago
Read 2 more answers
The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are
almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_3

So, the required probability = \frac{^{6}C_3}{^{8}C_3}  

                                              = \frac{20}{56}  = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_2

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_1

So, the required probability = \frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}  

                                                = \frac{30}{56}  = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = ^{8}C_3

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = ^{6}C_1

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = ^{2}C_2

So, the required probability = \frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}  

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(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0
3 years ago
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