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1 year ago

What is the mechanical advantage of a hockey stick that is gripped at point D and hits the puck at point B?

Mathematics

2 answers:

ladessa [460]1 year ago

7 0

Answer: Mechanical advantage = Effort distance/Resistance distance

Effort distance = D - Fulcrum = 0.5 m

Resistance distance = A - Fulcrum = 2.0 m

Then,

Mechanical advantage = 0.5/2 = 0.25

Hockey stick is an example of third class where mechanical advantage is less than one as the speed of the end of point A is critical.

Step-by-step explanation:

RideAnS [48]1 year ago

5 0

Answer:

Mechanical advantage = Effort distance/Resistance distance

Effort distance = D - Fulcrum = 0.5 m

Resistance distance = A - Fulcrum = 2.0 m

Then,

Mechanical advantage = 0.5/2 = 0.25

Hockey stick is an example of third class where mechanical advantage is less than one as the speed of the end of point A is critical.

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Select the expression that is equal to (-2x +9) – (3x – 4).

Anna [14]

Answer:

Step-by-step explanation:

(-2x+9-3x-4)

-2x-3x+9-4

-5x+5

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2 years ago

A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag

ahrayia [7]

Answer:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$ Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:

$X \sim N(M,4)$

Where $\mu=M$ and $\sigma=4$

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$

8 0

2 years ago

Choose a method for 16x60 using strategies!

OverLord2011 [107]

aline it like that then do 6×0 then put that answer ,you know what I'm jut going to show you. 60×

360

600

↑then add 360 and 600 and that's all. Sorry that the numbers look all funny.

6 0

2 years ago

The box plots below show the distribution of salaries, in thousands, among employees of two small companies.

Shtirlitz [24]

Answer:

Mean and IQR

Step-by-step explanation:

The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.

For company A:

Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.

Hence, the mean will be the best measure of centre rather Than the median in this case.

For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.

7 0

2 years ago

Read 2 more answers

If the length of segment RQ is 8, what is the length of segment PQ?

artcher [175]

Half of 8 is 4 so it’s b

7 0

2 years ago