Find the quotient. Simplify completely. Please show work.

Mathematics

1 answer:

nadya68 [22]3 years ago

8 0

1. 8/9 ÷ 4/9 is done by inverting the divisor (4/9) and then multiplying:

(8/9)(9/4) = 8/4 = 2 (answer)

2. 1/3 ÷ 6 is done similarly: (1/3)(1/6) = 1/18 (answer)

3. 4 ÷ 1/5 = 4 * 5 = 20 (answer)

4. 7 1/2 ÷ 1/4 = (15/2)(4/1) = 30 (answer)

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In a large population, 71 % of the people have been vaccinated. If 4 people are randomly selected, what is the probability that

Oksi-84 [34.3K]

Answer:

The answer is "0.9929"

Step-by-step explanation:

$\text{The distribution in binomials is n = 4 and p = 0.71.}\\\\\text{Vaccinated Pr(AT LEAST ONE) = 1-pr( no people vaccinated ).}$

$= 1 - Bin (n=4, k =0, p=0.71) \\\\= 1 - C(4,0) \times 0.71^0 \times (1-0.71)^{4-0} \\\\= 1 - (1 \times 0.29^4)$

$= 1 - (1 \times 0.00707281)\\\\=0.9929$

5 0

3 years ago

A runner warmed up for 10 minutes and then takes seven minutes to run each mile. The total time after r miles is 45 minutes how

scoundrel [369]

So our equation to find this can be represented by 7x + 10 = 45.
(x representing miles run)
Just solve for x!

Subtract the 10 from both sides, you've got
7x = 35
And now to isolate the x, we divide both sides by 7.
Now we're left with just
x = 5!

We can check this by substituting 5 as x in our equation.
7x x 5 + 10 = 45
45 = 45
It's right!

So the runner ran 5 miles total.

Hope this helps!
If I skimmed over this too much, let me know and I'll try to explain the best that I can.

7 0

3 years ago

Q1 How many subsets of a set with 100 elements have morethan one element ?

Damm [24]

A set of $n$ elements has $2^n$ possible subsets, where two classes of those sets are the empty set (1) and all the possible singleton sets ( $n$ ). So a set of $n$ elements has $2^n-1-n$ possible subsets with more than one elements. For Q1 take $n=100$ .

Q2a. Assuming not containing the same digits twice also includes not numbers with three or four of the same digit, and assuming digits are chosen from the usual 0-9, there are

$4!\dbinom{10}4=\dfrac{4!10!}{4!(10-4)!}=10\times9\times8\times7=5040$

possible strings.

Q2b. The first three digits can be chosen freely from 0-9, while the last digit has to be one of 0, 2, 4, 6, or 8. This means you have

$10^3\times5=5000$

possible strings

Q2c. Any such string will take the form $999X$ , $99X9$ , $9X99$ , or $X999$ , where $X$ has 9 possible choices (0-9 excluding 9, since we want exactly three 9s in any such string). So there are

$4\times9=36$

possible strings.

4 0

3 years ago

Can somebody evaluate this for me!! 8^4 Thank you!!

viktelen [127]

$\large\color{gold}{ \tt{8^{4}}}$