Ask question

Ask question

All categories

3 years ago

11

In a race in which twelve automobiles are entered and there are no ties, in how many ways can the first three finishers come i

n?

2 answers:

Ostrovityanka [42]3 years ago

5 0

Answer:

1320 ways

Step-by-step explanation:

In a race in which twelve automobiles are entered and there are no ties.

We have to find the ways in which first three finishers come in.

In these questions when order is fixed means when first, second and third positions are fixed (First comes, then second and then third) then we use the permutation to find the number of ways.

3 positions out of 12 automobiles = $^{12}P_{3}=\frac{12!}{(12-3)!}$

= $\frac{12!}{9!}$

= $\frac{(12)(11)(10)(9!)}{(9!)}$

= 12×11×10

= 1320

In 1320 ways first three automobiles may come in.

k0ka [10]3 years ago

3 0

That is 3 slots
12 in 1st slot
11 in 2nd
10 in 3rd

so that is 12*11*10 or 1320 ways

You might be interested in

70 POINTS PLEASE HELP!! Use the two equations: f (x) =x^2 and g (x) =4x+1 to determine the solutions to the following function o

bagirrra123 [75]

<h2><em><u>My</u></em><em><u> </u></em><em><u>equation</u></em><em><u> </u></em></h2>

<em>f</em><em>(</em><em>x</em><em>=</em><em>x</em><em>^</em><em>2</em><em> </em><em>is</em><em> </em><em>30</em><em> </em><em>and</em><em> </em><em>g</em><em>(</em><em>x</em><em>)</em><em> </em><em>=</em><em>4 1</em><em> </em><em> </em><em>is</em><em> </em><em>50</em><em>.</em><em>1</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>#brainliestbunch</em><em> </em><em>=</em><em>)</em><em> </em>

6 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

4 years ago

Is 3b-9 a algebraic expression

Slav-nsk [51]

Answer:

yes

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Tom swims a 1/2 kilometers every 1/4 hour. How far will he swim in one hour.

fgiga [73]

Answer:

2 kilometers

Step-by-step explanation:

every 1 kilometer is a 1/2 hour. double that and you get 2

8 0

3 years ago

Ron's recycle shop had an old paper-shredding machine. Business was good, so he bought a new paper-shredding machine. The old ma

levacccp [35]

Answer: 1 hour 20 minutes

Step-by-step explanation:

Given : The old machine could shred a truckload of paper in 4 hours.

Rate of work done by old machine = $\dfrac{1}{4}$

The new machine could shred the same truckload in 2 hours.

Rate of work done by new machine = $\dfrac{1}{2}$

Let 't' be the time taken by both of them working together , then we have the following equation:-

$\dfrac{1}{t}=\dfrac{1}{4}+\dfrac{1}{2}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{1+2}{4}\\\\\Rightarrow\dfrac{1}{t}=\dfrac{3}{4}\\\\\Rightarrow t=\dfrac{4}{3}=1\dfrac{1}{3}\ \text{ hours}$

Since 1 hour = 60 minutes

Then, $\dfrac{1}{3}\ \text{ hours}=\dfrac{1}{3}\times60=20\text{ minutes}$

Hence, it will take 1 hour 20 minutes to shred the same truckload of paper if Ron runs both shredders at the same time.

8 0

3 years ago