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scoundrel [369]
3 years ago
11

The price of a 36 pack of chips is $5.40, what is the price of one bag of chips?

Mathematics
2 answers:
Igoryamba3 years ago
5 0

Answer:

15 cents

Step-by-step explanation:

5.40/36 = .15

Rom4ik [11]3 years ago
4 0

Answer:

15 cents

Step-by-step explanation:

5.40÷36=15 cents

hope it helps

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A cone-shaped container has a height of 240 in. and a radius of 150 in. The cone is filled with a liquid chemical. The chemical
jeka57 [31]
It would take 471 weeks. First, square the radius. Then divide the height by three. Multiply the square radius and the answer you got after dividing h by 3. Then multiply that answer by 3.14. Usually, you're done, but for this problem take the extra step and divide it by 12,000.
5 0
3 years ago
I need help please I don’t know how to make this can someone help me ?
NISA [10]

Answer:

the simplified answer is 3 3/4

Step-by-step explanation:

8 0
3 years ago
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.

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

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
Is finding a car's depreciation over a 10 year period considered as univariate or bivariate?
g100num [7]

Determining a car's depreciation over a ten year period is considered a bivariate.

<h3>What is a bivariate?</h3>

A Bivariate data is made up of two variables that are observed against each other. In determining the deprecation of a car, the cost of the car is observed against the passage of time and the depreciation factor.  

To learn more about depreciation, please check: brainly.com/question/25552427

#SPJ1

8 0
2 years ago
Determine if the following system of equations has no solutions ......................
topjm [15]

Answer:

No solutions

Step-by-step explanation:

3x+4y=-4

15x+20y=-22

——————————

-5(3x+4y)=-5(-4) multiply 1st equation by -5 so you can eliminate the variables

-15x-20y=20. Modified 1st equation

15x+20y=-22.  2nd equation

0+0=-2. Added two equations together

since 0 does not equal -2, there is no solution.

8 0
3 years ago
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