What is the formula for area of a trapezuim??

What number should replace the letter x in the "Seed color" row?<br> 152<br> 207<br> 224<br> 2,001

Reptile [31]

Answer:c

Step-by-step explanation:

4 0

3 years ago

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 60 o

krok68 [10]

Answer:

a) 99.7% of the widget weights lie between 45 and 75 ounces

b) 97.2% of the widget weights lie between 50 and 75 ounces

c) 84% of the widget weights lie above 55

Step-by-step explanation:

The Empirical Rule states that:

50% of the values of a measure is above the mean, and the other 50% is below the mean.

99.7% of the values of a measure lie between 3 standard deviations of the mean.

95% of the values of a measure lie between 2 standard deviations of the mean.

68% of the values of a measure lie between 1 standard deviations of the mean.

In this problem, we have that: The widget weights have a mean of 60 ounces and a standard deviation of 5 ounces.

(a) 99.7% of the widget weights lie between

3 standard deviations of the mean, so:

60 - 3*5 = 60 + 3*5 = 45 and 75 ounces

(b) What percentage of the widget weights lie between 50 and 75 ounces?

We have to find the percentage that are below 75 and subtract by the percentage that are below 50. So

75 is 3 standard deviations above the mean. So 99.7% of the measures are below 75.

50 is 2 standard deviations below the mean. So only 5% of the measures that are below the mean are below 50.

So

99.7% - (50%)5% = 99.7% - 2.5% = 97.2%

(c) What percentage of the widget weights lie above 55?

55 is one standard deviation below the mean.

50% of the widget weights are above the mean.

Of the 50% that is below, 68% lie between one standard deviation(So from 55 to 60)

So

50% + 68%(50%) = 84%

3 0

3 years ago

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.