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3 years ago

What is it multiplying by?784.5, 78.45, 7.845,0.7845, 0.07845

Mathematics

1 answer:

katrin [286]3 years ago

3 0

Answer:

10,100,1000,-10,-100,-1000

Step-by-step explanation:

You might be interested in

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.08

kvv77 [185]

Answer:

a) 44.93% probability that there are no surface flaws in an auto's interior

b) 0.03% probability that none of the 10 cars has any surface flaws

c) 0.44% probability that at most 1 car has any surface flaws

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the binomial probability distributions.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson distribution with a mean of 0.08 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

So $\mu = 10*0.08 = 0.8$

(a) What is the probability that there are no surface flaws in an auto's interior?

Single car, so Poisson distribution. This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

44.93% probability that there are no surface flaws in an auto's interior

(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?

For each car, there is a $p = 0.4493$ probability of having no surface flaws. 10 cars, so n = 10. This is P(X = 10), binomial, since there are multiple cars and each of them has the same probability of not having a surface defect.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

0.03% probability that none of the 10 cars has any surface flaws

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

At least 9 cars without surface flaws. So

$P(X \geq 9) = P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{10,9}.(0.4493)^{9}.(0.5507)^{1} = 0.0041$

$P(X = 10) = C_{10,10}.(0.4493)^{10}.(0.5507)^{0} = 0.0003$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.0041 + 0.0003 = 0.0044$

0.44% probability that at most 1 car has any surface flaws

5 0

3 years ago

You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago

How many solutions does the equation 5x + 10 = 5x - 8 have?

andrew-mc [135]

Answer:

2 solutions

Step-by-step explanation:

5x+10=5x-8

5x-5x=8-10

=-2

5x+10=5x-8

10+8=5x-5x