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3 years ago

I need help on #9. The help would be grateful appreciated.

Mathematics

1 answer:

ELEN [110]3 years ago

8 0

Original Figure:
Length = 15
Width = 5
Height = 10
Volume = Length*Width*Height
Volume = 15*5*10
Volume = 750

New Figure
Length = 3
Width = 1
Height = 2
Each dimension has been divided by 5 (eg: 15/5 = 3)
Volume = Length*Width*Height
Volume = 3*1*2
Volume = 6

The old volume was 750 and it changes to 6
Notice how 750/6 = 125
Which can be rearranged to 750/125 = 6

Answer: if you divide the old volume by 125, then you get the new volume

Note: the new volume is 125 times smaller than the old volume
Put another way, the old volume is 125 times larger compared to the new volume

The fact that 125 = 5^3 is not a coincidence. If you divide each dimension by some number k, then you divide the volume by k^3

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A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys

Luba_88 [7]

Answer:

$z=-1.28$

And if we solve for a we got

$a=470 -1.28*60=393.2$

So the value of height that separates the bottom 10% (Fastest 10%) of data from the top 90% is 393.2.

so if the time is 393.2 or lower a boy would earn a certificate of recognition from the fitness association

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 time for this event of a population, and for this case we know the distribution for X is given by:

$X \sim N(470,60)$

Where $\mu=470$ and $\sigma=60$

NOTE: For this case the fastest 10% is on the lower tail of the distribution

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.9$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

$P(X$