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3 years ago

Solve 4a+3=11 plz step by step

2 answers:

6 0

4a + 3 = 11...subtract 3 from both sides
4a + 3 - 3 = 11 - 3...simplify
4a = 8 ....divide both sides by 4
(4/4)a = 8/4...simplify
1a, or just a = 2 <==

madam [21]3 years ago

6 0

Hi,

Again, we want to work on isolating the variable a. Let's start doing this by first getting rid of the 3.

We must do the inverse of the operation, which in this case will be subtraction since the 3 is positive. Also, always remember that when we do something on one side of the equation, we must do it on the other.

4a + 3 - 3 = 11 - 3

4a = 8

Since the 4 is multiplying our variable a, this means the inverse of the operation will be division. To fully isolate a by getting rid of the 4, we will divide by 4 on both sides.

4a / 4 = 8 / 4

a = 2

Hopefully, this 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?

1/3 probability that the first cabinet is chosen. This cabinet has two transistors, both of which are NPN, so 100% probability of selecting a NPN transistor.

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.

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

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

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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3 years ago

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Answer:

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$