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3 years ago

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

Mathematics

1 answer:

viva [34]3 years ago

8 0

Answer:

990 ways

Step-by-step explanation:

The total number of automobiles we have is 11.

Now, what this means is that for the first position , we shall be selecting 1 out of 11 automobiles, this can be done in 11 ways( 11C1 = 11!/(11-1)!1! = 11!/10!1! = 11 ways)

For the second position, since we have the first position already, the number of ways we can select the second position is selecting 1 out of available 10 and that can be done in 10 ways(10C1 ways = 10!9!1! = 10 ways)

For the third position, we have 9 automobiles and we want to select 1, this can be done in 9 ways(9C1 ways = 9!/8!1! = 9 ways)

Thus, the total number of ways the first three finishers come in = 11 * 10 * 9 = 990 ways

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What is the value of this expression when b = 5? 6(26-4)

romanna [79]

Answer:

6(26-4)= 132

Step-by-step explanation:

There isn't any variable in the equation

7 0

3 years ago

Read 2 more answers

Solve each inequality, and then drag the correct solution graph to the inequality.

Nesterboy [21]

The correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

$4(9x-18)>3(8x+12)$

First, clear the brackets,

$36x-72>24x+36$

Then, collect like terms

$36x-24x>36+72\\12x >108$

Now divide both sides by 12

$\frac{12x}{12} > \frac{108}{12}$

∴ $x > 9$

For the second inequality

$-\frac{1}{3}(12x+6) \geq -2x +14$

First, clear the fraction by multiplying both sides by 3

$3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)$

$-1(12x+6) \geq -6x +42$

Now, open the bracket

$-12x-6 \geq -6x +42$

Collect like terms

$-6 -42\geq -6x +12x$

$-48\geq 6x$

Divide both sides by 6

$\frac{-48}{6} \geq \frac{6x}{6}$

$-8\geq x$

∴ $x\leq -8$

For the third inequality,

$1.6(x+8)\geq 38.4$

First, clear the brackets

$1.6x + 12.8\geq 38.4$

Collect likes terms

$1.6x \geq 38.4-12.8$

$1.6x \geq 25.6$

Divide both sides by 1.6

$\frac{1.6x}{1.6}\geq \frac{25.6}{1.6}$

∴ $x \geq 16$

Let the graphs be A, B and C from left to right

The first graph (A) shows $x\leq -8$ and this matches the 2nd inequality

The second graph (B) shows $x \geq 16$ and this matches the 3rd inequality

The third graph (C) shows $x > 9$ and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

Learn more here: brainly.com/question/17448505

8 0

2 years ago

Seventh graders collected some gold sand on a class field trip.

satela [25.4K]

Answer:

Class 7A collected 4.8 ounces of pure gold.

Step-by-step explanation:

Key skills required are: Percentages, Multiplication

We only need the information about Class 7A. They collected 40 oz that contains 12% gold.
In other words, this means that 12% of that 40 oz gold sand is pure gold.

Here we have to do 12% x 40 to find the number of oz of pure gold. We first have to convert 12% into a decimal. Divide it by a 100 (or move the decimal point 2 places to the left) and you will get 0.12.

Do 0.12 x 40 and you will get 4.8

Therefore, there are 4.8 oz of pure gold in Class 7A's gold sand

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2 years ago

Consider the equation:

Alenkasestr [34]

Answer:

the equation is not correct....

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Sergeeva-Olga [200]

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3 years ago