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3 years ago

7

15=3y+7.y= please help me with my assignment i havelike 57 more questions to answer and 10% im trying by myself to solve the res

t are kinda difficult

1 answer:

monitta3 years ago

7 0

Answer:

8/3 or 2.67 to nearest hundredth.

Step-by-step explanation:

15 = 3y + 7

15 - 7 = 3y

3y = 8

y = 8/3

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What is the solution to the equation?

ki77a [65]

Answer:

7

Step-by-step explanation:

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3 years ago

Student finished eighth of homework problem in the class If the ratio of problem she finish to problem she still had left was 4.

omeli [17]

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3 years ago

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.

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3 years ago

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blondinia [14]

Answer:

I'm sorry for answering like that, so here is my actual answer: It is the third one. Again I'm very sorry.

Step-by-step explanation:

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2 years ago

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Is the answer B correct?

VashaNatasha [74]

If you do the elimination process, here it is.

- A : Too small

+ B : I think so

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- D : Way too big

7 0

3 years ago

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