Solve these..........

In testing the lethal concentration of a chemical found in polluted water. it is found that a certain concentration will kill 20

sasho [114]

Answer:

a) 0.1091

b) 0.9994

c) 0.5886

Step-by-step explanation:

X = the number of fish out of 20 that die after 24 hours

x = 0, 1, 2, . . . , 20

X~ Binomial (n= 20, p =0.20)

P(14 survive) = P(X = 6)

= $nC_r p^r q^{n-r}= 20C_6 0.2^6\times0.8^{14}$ =0.1091

Similarly we can find out

P(at least 10 survive) = P( X <= 10 ) = (Using technology) = 0.9994

P(at most 16 will survive) = P(X <= 16) = (Using technology) = 0.5886

7 0

3 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago

Which is the simplified form of the expression ((p squared) (q Superscript 5 Baseline)) Superscript negative 4 Baseline times ((

Alexus [3.1K]

Answer:

$q^{-30}$

Step-by-step explanation:

Given the expression:

$((p^2) (q ^5)) ^{-4} X ((p ^{-4}) (q^5)) ^{-2}$

First, we open the outer brackets

$=(p^{2X-4}) (q ^{5X-4})X (p ^{-4X-2}) (q^{5X-2})\\\\=(p^{-8}) (q ^{-20})X (p ^{8}) (q^{-10})\\\\$Next, we can collect like terms\\\\=p^{-8}Xp ^{8}Xq ^{-20}Xq ^{-10}\\\\$Applying addition law of indices\\\\=p^{-8+8}Xq^{-20+(-10)}\\\\=p^0 X q^{-30}\\\\=1 X q^{-30}\\\\=q^{-30}$

Therefore:

$((p^2) (q ^5)) ^{-4} X ((p ^{-4}) (q^5)) ^{-2}=q^{-30$

8 0

3 years ago

Read 2 more answers

You are training for a marathon. On the first three days of training, your run for 2,4, and 5 miles. How many miles should you r

jasenka [17]

The answer would have to be like the same as the first day, so that being said you would have to add. I think the answer is 2,2,4. hope this helps.

3 0

3 years ago

F(x)=- square root x-2. +3

sukhopar [10]

Answer:

Hi, The.

The domain of a function is the set of all possible values for x. The range is all possible values for y.

You should be able to see these pretty easily from the graph. If I understand your function correctly:

f(x) = square root(x+2) - 3

then x cannot be less than -2. Your domain is x > -2. In interval notation [-2,∞)

The lowest value for y is -3 so the range is y > -3. In interval notation [-3,∞)

Hope this helps!

7 0

3 years ago

Solve these..........​

Solve these..........