The answer would be C) because there are 11 possibilities of who can line up first, and after the first person, 10 possibilities, then 9, and so on. So you would need to multiply 11x10x9x8x...x1.

4 0

2 years ago

Help me please if you can

jolli1 [7]

It is either top left or bottom right, I cannot see the numbers clearly.

3 0

2 years ago

J & J’s Landscaping Co. makes about $320 a weekday, and ends up with weekly expenses of $1000. How much profit do they expec

guajiro [1.7K]

The answer is D. 320x 5 the. subtract 1000

8 0

1 year ago

Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the proba

Diano4ka-milaya [45]

Answer:

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Step-by-step explanation:

Let X = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = p = 0.05.

The sample of items selected is of size, n = 150.

The random variable X follows a Binomial distribution with parameters n = 150 and p = 0.05.

According to the Central limit theorem, if a sample of large size (n > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is: $\mu_{\hat p}= p=0.05$

The standard deviation of this sampling distribution is: $\sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178$

If 10 of the 150 items produced are unacceptable then the probability of this event is:

$\hat p=\frac{10}{150}=0.067$

Compute the value of $P(\hat p\leq 0.067)$ as follows:

$P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315$

*Use a z-table for the probability.

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

5 0

2 years ago