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4 years ago

I need help please fast

Mathematics

1 answer:

Slav-nsk [51]4 years ago

7 0

1. Subtract the largest number with the smallest number... e.g. 100 - 25

2. Divide the range by 5 and 10... e.g 75/10= 7.5 75/5=15

3. Pick the number between 5 and 10... e.g. 7.5

4. Round to the largest whole number... e.g. 8

5. Add 8 to the smallest number of the range 5 times... e.g. 25 + 8, 33 +8, etc

6. You'll want to have it so the previous number becomes your beginning of your intervals... e.g. 25-33, 33-41, 41-49, 49-57, 57-65.

6. Fill your histogram

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

Andrews [41]

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

6 0

3 years ago

What is the quotient ? -4/2 divided by 2

professor190 [17]

Answer:

$\frac{-\frac{4}{2} }{2} =-\frac{4}{2} *\frac{1}{2} =-\frac{4}{4} =-1$