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2 years ago

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A fence is guarding off a vegetable garden in the form of a rectangle. It has one side that is 10m greater than the other side.

Find the length of the fence needed to surround this garden if the area of the vegetable garden is 2184m2.

1 answer:

DanielleElmas [232]2 years ago

5 0

Answer:

The length of the fence needed to surround this garden is 188 meters.

Step-by-step explanation:

Given : A fence is guarding off a vegetable garden in the form of a rectangle. It has one side that is 10 m greater than the other side.

To find : The length of the fence needed to surround this garden if the area of the vegetable garden is 2184 m² ?

Solution :

Let the one side of rectangle be 'x'.

Then the other side is 'x+10'.

The area of the rectangle is 2184 m²,

i.e. $x(x+10)=2184$

$x^2+10x-2184=0$

Solve by middle term split,

$x^2+52x-42x-2184=0$

$x(x+52)-42(x+52)=0$

$(x+52)(x-42)=0$

$x=-52,42$

Reject negative value,

The side of the rectangle is 42 m.

The other side is 42+10=52 m

The perimeter of the rectangle is $P=2(l+b)$

$P=2(42+52)$

$P=2(94)$

$P=188$

Therefore, the length of the fence needed to surround this garden is 188 meter.

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Step-by-step explanation:

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3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

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Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

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d) $z=1.28$

And if we solve for a we got

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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".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

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

$P(X>a)=0.1$ (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 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

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

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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Step-by-step explanation:

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