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padilas [110]
2 years ago
13

What’s 8765 times 9876

Mathematics
2 answers:
WITCHER [35]2 years ago
8 0
86,563,140 is the answer
Anastasy [175]2 years ago
4 0
Its 86,563,140 and how do you add a question
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40 is 60% of what number?
11111nata11111 [884]
40(60/100) 4(6/10) =24/10 = 2.4
4 0
2 years ago
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Given the function f(x) = 2(x + 6), find x if f(x) = 22.
7nadin3 [17]
Hello,

2(x+6)=22
==>x+6=11 divide by 2
==>x=5

6 0
2 years ago
Mustafa’s horse ate 6,400 pounds of hay last year. How could Mustafa determine the number of tons of hay his horse ate last year
svet-max [94.6K]

Answer:

C

Step-by-step explanation:

A ton is 2000 pounds, and in order to convert to a larger unit you must divide.

3 0
2 years ago
Brad has 17 balloons. 8 balloons are red and the rest are green. How many green balloons does Brad have?
Igoryamba

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9 green balloons

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2 years ago
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You are given a black box f : Z10 → Z10 that contains either a random permutation or a random function. Your distinguisher is al
SashulF [63]

Solution :

The function :    $f: Z_{10} \rightarrow Z_{10}$   be a random permutation.

f is a permutation on $Z_{10}$ ,  i.e. f is permutation 10.

Now we know that the total number of distinct permutation on to symbolize 10!.

Each of these 10! permutation to a permutation function  $f: Z_{10} \rightarrow Z_{10}$  

Therefore, total number of permutation functions  $f: Z_{10} \rightarrow Z_{10}$   are 10!.

Now we want the total number of permutation functioning :

$f: Z_{10} \rightarrow Z_{10}$   such that  f(0) = 0 and f(1)= 1

Now we notice that when f(0)=0 and f(1)=1, then two symbol '0' and '1 are fixed under permutation f.

So essentially when f(0) = 0 and f(1) = 1, f becomes permutation on 8 symbol.

Total number of permutation functioning  $f: Z_{10} \rightarrow Z_{10}$  , f(0)=0 and f(1)=1 are 8!

Now we want the probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0) = 0 and f(1) = 1.

The number of permutation function  $f: Z_{10} \rightarrow Z_{10}$  , i.e.

The probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0) = 0 and f(1) = 1 is

$\frac{8!}{10!} = \frac{8!}{10 \times 9\times 8!} =\frac{1}{10 \times 9}=\frac{1}{90}$

Therefore, the probability that a random permutation  $f: Z_{10} \rightarrow Z_{10}$   satisfies f(0)= 0 and f(1)=1 is  $\frac{1}{90}$

6 0
2 years ago
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