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nirvana33 [79]
3 years ago
13

How do you figure out the value of 1/10th of 20.44

Mathematics
1 answer:
KatRina [158]3 years ago
3 0
Hedge ghettos was the best way for me today was gonna you wanna come back home from home today for a
You might be interested in
How many numbers greater than zero and less than 50 equal the product of two consecutive positive integers?
spin [16.1K]
Consecutive numbers would be like 2 and 3, or 7 and 8.

All we need to do is keep multiplying pairs of consecutive numbers until we get above 50.

1 × 2 = 2 (that's one.)
2 × 3 = 6 (two)
3 × 4 = 12 (three)
4 × 5 = 20 (four)
5 × 6 = 30 (five)
6 × 7 = 42 (six...)
<em>7 × 8 = 56 > 50</em>
We have a total of 6 numbers that equal the product of 2 consecutive intergers<em>

</em>
7 0
2 years ago
What happens when warm air collides with cold air and is forced to rise over cold dome.
igor_vitrenko [27]

Answer:

Something to raise the moist air to form the clouds and cause precipitation. An example of lift is warm air colliding with cold air and being forced to rise over the cold dome. The boundary between the warm and cold air masses is called a front.

Step-by-step explanation:

6 0
2 years ago
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
2 years ago
How do you Simplify (y^12)^5
goldenfox [79]

Answer:

y^60

Step-by-step explanation:

(a^b)^b= a ^ b times b

12 times 5 is 60 so it's y^60

sorry my explaination is so complicated but im confident in the answer

8 0
2 years ago
Read 2 more answers
Can anyone help me out
Genrish500 [490]

Answer:

12k^5 +24k⁴+30k³

Step-by-step explanation:

the area = 6k³(2k²+4k+5)

= 12k^5 +24k⁴+30k³

5 0
3 years ago
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