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Marianna [84]
3 years ago
13

The number 6√3 is between what two integers and is closer to which integer on the number line.

Mathematics
1 answer:
UNO [17]3 years ago
5 0
Between 6&7 is closer to 10
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5(x−3.8)=22 <br> please help find the x value
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x = 8.2

Step-by-step explanation:

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At​ most, how many one sixth 1 6​-inch-long pieces of string can you cut from a two thirds 2 3​-inch-long piece of​ string?
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2\3 23-1\6 16
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Jack mails 10 packages that each weigh the same amount. If the combined weigh of all 10 package is 67 1/2 pounds, how much does
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3 years ago
14(0.89)^x+4.5 =162
lisabon 2012 [21]
14(0.89)^{x + 4.5} = 162
\frac{14(0.89)^{x + 4.5}}{14} = \frac{162}{14}
0.89^{x + 4.5} = 11\frac{4}{7}
ln(0.89^{x + 4.5}) = ln(11\frac{4}{7})
(x + 4.5)ln(0.89) = ln(11\frac{4}{7})
\frac{(x + 4.5)ln(0.89)}{ln(0.89)} = \frac{ln(11\frac{4}{7})}{ln(0.89)}
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3 0
3 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
3 years ago
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