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Solnce55 [7]
3 years ago
15

2 + 8 divided by 2 x 4

Mathematics
2 answers:
Katena32 [7]3 years ago
8 0
2+8 divided by 2x 4 is 17
yaroslaw [1]3 years ago
5 0
2 + 8 divided by 2 x 4 is 1 remainder 2
2 + 8 = 10
2 x 4 = 8
10 divided by 8 is 1 remainder 2
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there are between 24 to 40 students in a class. the ratio of boys to girls is 4:7. how many students are in the class.
ZanzabumX [31]

Answer:

8 students

Step-by-step explanation:

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Drupady [299]

Answer:

  • (a) 1/2
  • (b) 1/3

Step-by-step explanation:

(a) 3 of the 6 numbers on the die are even, so the probability that one of them will show is 3/6 = 1/2.

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(b) 2 of the 6 numbers on the die are less than 3, so the probability that one of them will show is 2/6 = 1/3.

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Find the equation of the line passing through the points (0,1) and (-2,0). write the equation in slope-intercept form.
Misha Larkins [42]

First of all, you have to calculate slope of that,

m = (y2-y1)/(x2-x1) = (1-0)/(0+2) = 1/2

now, equation would be y-y1 = m(x-x1)

y-1 = 1/2(x-0)

y-1 = x/2

y = x/2+ 1

5 0
3 years ago
During the day, Sam spent $4.85 on lunch. He also bought two books for $7.95 each at the end of the day he had $8.20 left.How mu
ch4aika [34]

We can add the money he spent to the money he had left to find the total.

4.85 + 15.9 + 8.20 = 28.95


He started with 28.95

6 0
3 years ago
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A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates
KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}     (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0
3 years ago
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