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spayn [35]
3 years ago
13

(this is multiple choice and i raised points make sure to EXPLAIN !)

Mathematics
2 answers:
otez555 [7]3 years ago
6 0

Answer:

C

Step-by-step explanation:

Basically (f+g)(x) = f(x)+g(x).

Following this rule you combine like terms.

7x-2x=5x and 10+5 = 15.

Therefore the answer is C.

Anna [14]3 years ago
6 0

f(x) = 10 - 2x

g(x) = 7x + 5

(f + g) (x) = (10 - 2x) + (7x + 5) //Adding two functions

= 10 - 2x + 7x + 5

= 15 + 5x

Answer: C.

//Hope it helps.

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An electronics hobbyist has three electronic parts cabinets with two drawers each.
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Answer:

a) 0.5 = 50% probability that an NPN transistor will be selected.

b) 0.3333 = 33.33% probability that it came from the cabinet that contains both types

c) 66.67% probability that it comes from the cabinet that contains only NPN transistors

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

a) What is the probability that an NPN transistor will be selected?

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1/3 probability that the second cabinet is chosen. This cabinet has two transistors, both of which are PNP, so 0% probability of selecting a NPN transistor.

1/3 probability that the second cabinet is chosen. This cabinet has two transistors, one of which is NPN, so 50% probability of selecting a NPN transistor.

So

p = \frac{1}{3}*1 + \frac{1}{3}*0 + \frac{1}{3}*0.5 = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = 0.5

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b) Given that the hobbyist selects an NPN transistor, what is the probability that it came from the cabinet that contains both types?

Here we use the conditional probability formula.

Event A: NPN transistor

Event B: From the third cabinet.

50% probability that an NPN transistor will be selected, so P(A) = 0.5.

1/6 probability that it is from the third cabinet and NPN, so P(A \cap B) = \frac{1}{6}

The desired probability is:

P(B|A) = \frac{\frac{1}{6}}{0.5} = 0.3333

0.3333 = 33.33% probability that it came from the cabinet that contains both types.

c) Given that an NPN transistor is selected what is the probability that it comes from the cabinet that contains only NPN transistors?

Either it comes from the cabinet with only NPN transistors, or it comes from the cabinet with both types of transistors. The sum of the probabilities of these outcomes is 100%. So

x + 33.33 = 100

x = 66.67

66.67% probability that it comes from the cabinet that contains only NPN transistors

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