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avanturin [10]
2 years ago
4

What is 224.7 - 58.646= ?​

Mathematics
2 answers:
riadik2000 [5.3K]2 years ago
6 0

Answer:

166.054

Step-by-step explanation:

Zanzabum2 years ago
6 0

Answer:

166.054

Step-by-step explanation:

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Four men are to divide K500 equally among them. When the money was given, 20% was taken away.
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Answer: 20% of 500= 100

So 500-100 = 400

4x100= 400

Step-by-step explanation:

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You bought $2000 worth of stocks in 2012. The value of the stocks has been decreasing by 10% each year. What will your stock be
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Answer:

that is the solution to the question

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An exterior angle of an isosceles triangle has measure 130°. Find two possible sets of measures for the angles of the triangle.
frozen [14]
(1) it can be both exterior of vertex or base. 180-130=50°
vertex is 50°:     base=(180-50)/2=65°
base is 50°:       vertex= 180-2*50=80°

vertex 50 base 65; vertex 80, base 50

(2)base=180-130=50°
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base 50 vertex 80
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Stephen spent $4 on milk, $6 on eggs, and $11 on cereal. He wrote the ratio 6/11 to describe some of his purchases. Explain why
o-na [289]
Attached the solution and work.

6 0
2 years ago
Read 2 more answers
A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe
xxMikexx [17]

Answer:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

Step-by-step explanation:

For the random variable N_1 we define the possible values for this variable on this case [1,2,3,4,5] . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable N_1 like this:

P(N_1 = a) = \frac{5-a C 1}{5C2}

For the distribution of N_2 we need to take in count that we are finding a conditional distribution. N_2 given N_1 =a, for this case we see that N_2 \in [1,2,...,5-a], so then exist 5-a C 1 ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

P(N_2 =b | N_1 = a) = \frac{1}{5-a C 1}

And if we want to find the joint probability we just need to do this:

P(N_1 = a , N_2 = b) = P(N_2 = b | N_1 = a) P(N_1 =a)

And if we multiply the probabilities founded we got:

P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}

8 0
2 years ago
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