<h3>5 of the 6 vaccinated people will be protected from the virus</h3>
<em><u>Solution:</u></em>
Given that,
Suppose that a vaccine is 85% effective against the flu
You vaccinated 6 people
To find: Number of people you expect will be protected from the virus
From given,
85 % of 6 people are effective against the flu
Which means,
![\rightarrow 85 \% \text{ of } 6\\\\\rightarrow \frac{85}{100} \times 6\\\\\rightarrow 0.85 \times 6\\\\\rightarrow 5.1 \approx 5](https://tex.z-dn.net/?f=%5Crightarrow%2085%20%5C%25%20%5Ctext%7B%20of%20%7D%206%5C%5C%5C%5C%5Crightarrow%20%5Cfrac%7B85%7D%7B100%7D%20%5Ctimes%206%5C%5C%5C%5C%5Crightarrow%200.85%20%5Ctimes%206%5C%5C%5C%5C%5Crightarrow%205.1%20%5Capprox%205)
Thus, 5 of the 6 vaccinated people will be protected from the virus
Answer:
4
Step-by-step explanation:
20%=0.2
0.2 times 20 will give you the answer 4
5 5/16 is greater.
You can find that out by dividing 5/16 and 7/24
So that gives .31 and .29
Thus, 5 5/16 are greater
1) find the mean,
20+10+30+20+50+30+40+20/8= 27.5
2) Find the distance of each value from that mean
20- 7.5
10- 17.5
30- 2.5
20-7.5
50- 22.5
30- 2.5
40- 12.5
20- 7.5
3) Find the mean of those distances
7.5+17.5+2.5+7.5+22.5+2.5+12.5+7.5/8=10
The Mean absolute deviation for this data set is 10.
Foil (x-1)(x-1)(x-1) gives you x^3-3x^2+3x+1
(<span>x^3-3x^2+3x+1)(3X-3) foil
</span>gives you 2x^4-9x^3+15x^2+11x-3
for the other side
(x-1)(x-1)= x^2-2x+1
(2x+1)(x^2-2x+1) = 2x^3-3x^2+1
(2x^4-9x^3+15x^2+11x-3) - (<span>2x^3-3x^2+1)
</span>=2x^4-11x^3+18x^2+11x-4