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3 years ago

Ummm how do you do this two questions?

Mathematics

1 answer:

guajiro [1.7K]3 years ago

8 0

Answer:

Here are the formulae you need. Just plug in the values you have and you will get your answers.

Area of a triangle: 1/2*base*height

Volume of a rectangular prism: length*width*height

Volume of a cylinder: pi*r^2*height

Step-by-step explanation:

You might be interested in

The number of typing errors made by a typist has a Poisson distribution with an average of three errors per page. If more than t

damaskus [11]

Answer:

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

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.

Poisson distribution with an average of three errors per page

This means that $\mu = 3$

What is the probability that a randomly selected page does not need to be retyped?

Probability of at most 3 errors, so:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0498 + 0.1494 + 0.2240 + 0.2240 = 0.6472$

0.6472 = 64.72% probability that a randomly selected page does not need to be retyped.

3 0

3 years ago

Graph: g(x)=5cos((\pi )/(2)x-(3\pi )/(2))-2

Tatiana [17]

$g(x)=5cos((\pi )/(2)x-(3\pi )/(2))-2$

generate by: Amplitude:5 Period:4

Phase shift:(3 to the right) Vertical shift:-2

x=3,g(x)= 3

x=4,g(x)= -2

x=5,g(x)= -5

x=6,g(x)= -2

x=7,g(x)= 3

the graph is like cos(x)

learn more about trigonometric graphs here:

brainly.com/question/18265536

#SPJ1

3 0

1 year ago

It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of responde

vivado [14]

Answer:

0.84 or 84%

Step-by-step explanation:

Let 'n' be number of people who took the survey, the number of men who answered the survey is 0.45n. The number of men who answered the survey and have participated in online auctions is 0.38n. Therefore, choosing from the population of men who took the survey (0.45n) the probability that a respondent had participated in an online auction is:

$P =\frac{0.38n}{0.45n}\\P=0.84$

8 0

3 years ago

What type of number is 0?

FrozenT [24]

Answer:

0 is a rational number, whole number, integer, and a real number.

Step-by-step explanation:

7 0

3 years ago

Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule

Vaselesa [24]

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

a) What is the approximate percentage of men between 169 and 183 cm?

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

$\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183$

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

b) Between which 2 heights would 95% of men fall?

This corresponds to ±2 standard deviations off the mean.

$\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190$

95% of the men will fall between 162 cm and 190 cm.

c) Is it unusual for a man to be more than 197 cm tall?

The number of standard deviations of distance from the mean is

$n=(197-176)/7=3$

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

3 0

3 years ago