3 years ago

Plz answer correctly. its for a test

Mathematics

1 answer:

4vir4ik [10]3 years ago

5 0

Answer:

the third one

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Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi

Sunny_sXe [5.5K]

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.

5 0

3 years ago

Ill mark brainist plss help

worty [1.4K]

Answer:

3.3 cm

Step-by-step explanation:

262/80 = 3.275 ≈ 3.3

3 0

3 years ago

Nth Term help please

liberstina [14]

$\bf \begin{array}{|ll|ll} \cline{1-2} term&value\\ \cline{1-2} a_1&3\\[0.8em] a_2&\stackrel{3\cdot 2}{6}\\[0.8em] a_3&\stackrel{6\cdot 2}{12}\\[0.8em] a_4&\stackrel{12\cdot 2}{24}\\[0.8em] a_5&\stackrel{24\cdot 2}{48}\\[0.8em] a_6&\stackrel{48\cdot 2}{96} \\ \cline{1-2} \end{array}$