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2 years ago

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The owner of a used car lot wants to buy many cars at an auction as he can while not spending more than $20000. If he expects th

e average price to be $4000, how many would he expect to be able to buy?

1 answer:

AleksAgata [21]2 years ago

7 0

Answer:

5

Step-by-step explanation:

Given that:

Total maximum amount that the owner wishes to spend = $20000

Average price of each car = $4000

To find:

How many cars that the owner can expect to buy?

Solution:

Total number of cars that the owner can expect to buy can be found by dividing the total money available with the owner with the average price of each car.

i.e.

$\text{Number of cars expected to be bought} = \dfrac{\text{Total money available with the owner}}{\text{Average Price of car}}$

We have the following values as given in the question statement:

Total money available = $20000

Average price of car = $4000

Therefore, the answer is:

$\text{Number of cars expected to be bought} = \dfrac{20000}{4000} = \bold{5}$

The owner can expect to buy 5 number of cars.

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3 years ago

Lily has a bottle containing 7/8 quart of milk. She pours 4/5 of it into a bowl. What amount of milk does she pour into the bowl

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2 years ago

Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

3 0

3 years ago

Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46

JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

$M = \frac{50463-38737}{2} = 5863$

Relating the margin of error with the sample size:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.64.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this problem, we have that $\sigma = 14300, M = 5863$ . So

$M = z\frac{\sigma}{\sqrt{n}}$

$5863 = 1.645\frac{14300}{\sqrt{n}}$

$5863\sqrt{n} = 1.645*14300$

$\sqrt{n} = \frac{1.645*14300}{5863}$

$(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2$

$n = 16$

The sample size was of 16.

7 0

3 years ago