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2 years ago

What is 34% of 60000000

Mathematics

2 answers:

Ainat [17]2 years ago

7 0

(Lets hope i'm right :p)

The first thing you do is make "34%" into a fraction. That would be 1/34.

After, you would multiply.

1/4 x 60000000/1 = ?

(Use cross multiply if you know)

60000000 divided by 4 = 15000000

1/1 x 15000000/1 = 15000000/1 OR 15000000

Answer: 15000000/1 or 15000000

<em>Tell me if i'm wrong so I can prevent this in the future! Have a lovely day!</em>

~Pooch ♥

mafiozo [28]2 years ago

6 0

20400000 is the answer

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Based only on the information given in the diagram it is guaranteed that RST UVW

lapo4ka [179]

Answer:

The triangles are similar

Step-by-step explanation:

We are given and two angles so by default we know the third angle. We know the three angles are the same, so the triangles are similar by AAA. We do not about conqruency without a side length.

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2 years ago

Read 3 more answers

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

2 years ago

Two step equation -5=d/4+3

Karolina [17]

Answer:

If you are looking for d the answer is -32.

Step-by-step explanation:

-5=d/4+3

-3 -3

(4)-8=d/4(4)

-32 =d

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1 year ago

I need help with this problem

alexandr402 [8]

Answer:

4*x^4*y^22

Step-by-step explanation:

Your goal here is to REDUCE the given expression to simplest terms.

One way in which to approach this problem would be to rewrite (2x^2y^10)^3 as: (2x^2*y^8)*y^2*(2x^2*y^10)^2.

Dividing this rewritten expression by 2x^2*y^8 results in:

y^2(2x^2*y^10)^2.

We now need to raise (2x^2*y^10) to the power 2. Doing this, we get:

4x^4*y^20.

Multiply this by y^2 (see above):

y^2*4x^4*y^20

The first factor is 4: 4y^2*x^4*y^20. This is followed by the product of y^2 and y^20: 4*y^22*x^4

Finally, this should be re-written as

4*x^4*y^22

Another way of doing this problem would involve expanding the numerator fully and then cancelling out like factors:

8*x^6*y^30 4*x^4*y^22

----------------- = ------------------ = 4*x^4*y^22

2x^2y^8 1

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2 years ago

∆ABC is isosceles, AB = BC, and CH is an altitude. How long is AC, if CH = 84 cm and m∠HBC = m∠BAC +m∠BCH?

son4ous [18]

I think I could help you
Have fun and feel free to ask me something new.
Or we can prove some properities without calculating by details

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2 years ago