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NikAS [45]
3 years ago
8

Amy is doing a science experiment on how a certain bacterium reacts to an antibiotic. She has 3 dishes of identical bacterium sa

mples with 14 bacteria in each dish. She gives an antibiotic to all of the bacteria in one dish. All of the treated bacteria died, and the bacteria in the other two dishes survived.
What type of sampling is demonstrated in the situation above?
negative sampling
convenience sampling
random sampling
census
Mathematics
1 answer:
Katarina [22]3 years ago
5 0

ANSWER EQUALS 567. I THINK


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daser333 [38]
N(N ∩ S ∩ K) = 10
n(ξ) = 250
n(S ∪ K) = 15 - 10 = 5
n(N ∪ S) = 20 - 10 = 10
n(N ∪ K) = 30 - 10 = 20
n(S) = 50 - 10 - 5 - 10 = 25
n(K) = 55 - 20 - 5 - 10 = 20
n(N) = 100 - 10 - 20 - 10 = 60

n(N ∪ S ∪ K) = 10 + 5 + 10 + 20 + 25 + 20 + 60 = 150

Therefore, n(N ∪ S ∪ K)' = 250 - 150 = 100

Therefore, 100 million people do not read any of the three papers.
7 0
3 years ago
Jason left a bin outside in his garden to collect rainwater. He notices that 1 over 5 gallon of water fills 2 over 3 of the bin.
sergiy2304 [10]

Answer: 3/10

Step-by-step explanation: (1/5) gallon / (2/3) bin = x gallons / 1 bin cross-multiply

1/5 = (2/3)x multiply both sides by the reciprocal of 2/3 = 3/2

(3/2) (1/5) = x = 3/10 gallons fills the entire bin

Look at it this way......each 1/10 of a gallon fills 1/3 of the bin......so 3(1/10) gallons fills 3(1/3)bins....or

3/10 gallons fills a whole bin...

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2 years ago
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Helpp plzzz I will mark brainliestttt
miskamm [114]

Answer:

it 152.3

Step-by-step explanation:

6 0
3 years ago
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Why did i get this wrong? What did i have to do to get the right answer?
aalyn [17]
Will u did your math wong 
6 0
3 years ago
"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx

For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)

So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
7 0
3 years ago
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