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3 years ago

I need help with this does anyone understand this?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

Let y = f(x)

$= > y = \frac{19}{ {x}^{3} }$

$= > {x}^{3} = \frac{19}{y}$

$= > x = \sqrt[ 3]{ \frac{19}{y} }$

${f}^{ - 1} (x) = \frac{ \sqrt[3]{19} }{ \sqrt[3]{x} }$

<h2>option A</h2>

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John Glenn High School has approximately 1000 students. A survey randomly asked 20 students if they regularly buy lunch. Of thos

lutik1710 [3]

Answer:

Here we have the table for the 8 students that said yes.

a) Out of the 8 students, 5 purchased their favorite food more than four times a month.

The proportion will be p = 5/8, and this can not be simplified anymore because 5 is a prime number and 5 is not a factor of 8.

b) The mean time will be equal to the amount of each of the 8 purchased their favourite food divided by the total sample (8)

This is:

m = (6 + 4 + 3 + 8 +6 + 9 + 4 + 7)/8 = 5.875

c) Out of a sample of 20, 8 buy lunch regularly, this is; p1 = 8/20

The amount of students that regularly buy food will be:

p1*1000 = (8/20)*1000 = 400

And of those 400, 5/8 buy their favourite lunch more than 4 times per month:

(5/8)*400 = 250

d: The mean number of times that a student regularly buys salad.

We have 3 students that regularly buy salad (so we only do the math with those 3)

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3 years ago

If 2a-8=4 and 3b-8=10, what is the value of a+b?

Irina-Kira [14]

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2 years ago

Read 2 more answers

Say a hacker has a list of n distinct password candidates, only one of which will successfully log her into a secure system. a.

alexdok [17]

Answer:

The probability is $\frac{1}{n}$

Step-by-step explanation:

If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:

If k=1

$P = \frac{1}{n}$

Because, in her first try she has n possibles options and just one give her a successful login.

If k=2

$P=\frac{n-1}{n} *\frac{1}{n-1} =\frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.

If k=3

$P=\frac{n-1}{n} *\frac{n-2}{n-1} *\frac{1}{n-2} = \frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.

Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n

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