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torisob [31]
3 years ago
10

If a consistent system of equations has more unknowns than equations, what can be said about the number of solutions?

Mathematics
1 answer:
Darya [45]3 years ago
4 0
Then there are not enough equations in order to solve the system.
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What is thw answer to x/5=2
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pamela wants to buy a new mercedes.The mercedes wheel has a radius of 35', what is the area of the mercedes wheel?
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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}  

                                                = \frac{6}{56}  = <u>0.11</u>

(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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