I NEED HELP !! Answer to this?

Suppose "35" cars start at a car race. In how many ways can the top 3 cars finish the race?

Thepotemich [5.8K]

Answer:

The value is $\left 35 } \atop }} \right. P_3 = 39270$

Step-by-step explanation:

From the question we are told that

The total number of cars is $n = 35$

The number cars considered is $r = 3$

Generally the number of different top three finishes possible for this race of 35 cars is mathematically represented as

$\left n } \atop }} \right. P_r = \frac{n!}{(n - r) !}$

$\left 35 } \atop }} \right. P_3 = \frac{35! }{(35 - 3) !}$

$\left 35 } \atop }} \right. P_3 = \frac{35 * 34 * 33 * 32! }{32 !}$

$\left 35 } \atop }} \right. P_3 = 39270$

7 0

2 years ago

Happy Monday Everybody Free 100 points no joke

kirill [66]

Answer:

YESSS cheers mate

7 0

3 years ago

Read 2 more answers

Write the numerical expression 34 in expanded form.

garik1379 [7]

Answer:

not sure what type you mean but here are options

Step-by-step explanation:

30+4=34

thirty-four

5 0

2 years ago

How do I do this qn?

denis-greek [22]

You can take the log of the left and right hand side, and then apply the logarithm rules:

log(aˣ) = x·log(a)
log(ab) = log(a) + log(b)

log(9^(x-1) * 2^(2x+2)) = log(6^(3x))
log(9^(x-1)) + log(2^(2x+2)) = 3x log(6)
(x-1) log(9) + (2x+2) log(2) - 3x log(6) = 0
x(log9 + 2log2 - 3log6) = log9 - 2log2
x = (log9 - 2log2) / (log9 + 2log2 - 3log6)

simplifying by writing log9 = 2log3 and log6 = log2+log3

x= 2(log3 - log2) / (2log3 + 2log2 - 3log2 - 3log3) =
x= -2(log3 - log2) / (log3 + log2) = -2 log(3/2) / log(6)

So 6^x = 4/9

3 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$