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Sergio [31]
3 years ago
9

What is 2 3/12 minus 1 8/12 equal to?​

Mathematics
2 answers:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

7/12

Step-by-step explanation:

3/12=1/4

2 1/4=9/4

8/12=2/3

1 2/3=5/3

---------------

9/4-5/3=27/12-20/12=7/12

zavuch27 [327]3 years ago
3 0

Answer:

1.5

Step-by-step explanation:

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What is the answer to 0.5(2x + 8) = x - 4
svp [43]

0.5(2x+8) = x-4\\\\\implies \dfrac 12 (2x+8 ) = x-4\\\\\implies  2x+8 =  2x-8\\\\\implies 8=-8\\\\\text{Which is not true, so there are no solution.}

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2 years ago
Which graph best represents the equation x2 + (y – 1)2 = 3?
Harman [31]

Answer:

Step-by-step explanation:

assuming the equation is x^2 + (y - 1)^2 = 3

6 0
2 years ago
The​ life, in​ years, of a certain type of electrical switch has an exponential distribution with an average life β=44. If 100 o
Bond [772]

Answer:

0.9999

Step-by-step explanation:

Let X be the random variable that measures the time that a switch will survive.

If X has an exponential distribution with an average life β=44, then the probability that a switch will survive less than n years is given by

\bf P(X

So, the probability that a switch fails in the first year is

\bf P(X

Now we have 100 of these switches installed in different systems, and let Y be the random variable that measures the the probability that exactly k switches will fail in the first year.

Y can be modeled with a binomial distribution where the probability of “success” (failure of a switch) equals 0.0225 and  

\bf P(Y=k)=\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}

where  

\bf \binom{100}{k} equals combinations of 100 taken k at a time.

The probability that at most 15 fail during the first year is

\bf \sum_{k=0}^{15}\binom{100}{k}(0.02247)^k(1-0.02247)^{100-k}=0.9999

3 0
3 years ago
A student takes a driving test until it is passed.
Leya [2.2K]

Answer:

The probability that the test is taken an even number of times is 0.30.

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

P (even number of tests) = P (Passing in the 2nd attempt)

                                                  + P (Passing in the 4th attempt)

                                                       + P (Passing in the 6th attempt) ...

If a student passed in the 2nd attempt it implies that he failed in the first.

Then,  P (Passing in the 2nd attempt) = (\frac{3}{7}) \times (\frac{4}{7})

Similarly, P (Passing in the 4th attempt) = (\frac{3}{7})^{3} \times (\frac{4}{7}), since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = (\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

S=\frac{a}{1-r^{2}}

The probability is:

P (even number of tests) = (\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...

                                          =\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\  =\frac{12}{40}\\ =0.30

Thus, the probability that the test is taken an even number of times is 0.30.

7 0
3 years ago
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