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chubhunter [2.5K]
3 years ago
11

A student wrote a number pattern such that each number is 3 more that 4 times the previous number. What is the sixth term in the

pattern if the first term is 6?
Mathematics
2 answers:
Serga [27]3 years ago
8 0
1st term = 6
2nd term = 24+ 3
3nd term = 111
4th term = 447
5th term = 1791
6th term = 7167

Is there a simplified formula?
Nesterboy [21]3 years ago
4 0
Start with 1.
1st number = 1
2nd number = 1 * 4 + 3 = 7
3rd number = 7 * 4 + 3 = 31
4th = 31 * 4 + 3 = 127
5th = 127 * 4 + 3 = 511
Which means the 6th = 511 * 4 + 3, which is 2047.
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Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

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3 years ago
I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value o
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The expected values from the product of 2 marbles are 1,2,3,4,5,6,8,10,12,15,20

Step-by-step explanation:

THe expected values are 1,2,3,4,5,6,8,10,12,15,20

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