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3 years ago

PLEASE GUYS HELP 60 POINTS PLS HELP

Mathematics

1 answer:

leonid [27]3 years ago

7 0

Answer:

$e\frac{4}{3}$ will be the exponential expression.

Step-by-step explanation:

We have to choose a number between 0 ad 10 and has to be odd to be base say 3

And even number between 0 ad 10 to be exponent say 4.

So the exponential expression will be : $e\frac{4}{3}$

According to the information given above expression is the required expression.

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GIVING AWAY 100 POINTS!!! BE THE FIRST TWO

Lelu [443]

Answer:

Thank you!!!

Step-by-step explanation:

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2 years ago

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A used car dealer has 30 cars and 10 of them are lemons (i.e.~ mechanically faulty used cars), and you don't know which is which

Triss [41]

Answer: Hence, the probability that he will get at least one lemon is 0.70.

Step-by-step explanation:

Since we have given that

Number of cars = 30

Number of lemon cars = 10

Number of other than lemon cars = 30-10 = 20

According to question, he bought 3 cars,

we need to find the probability that you will get at least one lemon.

So, P(X≤1)=1-P(X=0)=1-P(no lemon)

Here, P(no lemon ) is given by

$\dfrac{20}{30}\times \dfrac{20}{30}\times \dfrac{20}{30}=(\dfrac{20}{30})^3$

so, it becomes,

$P(X\geq 1)=1-(\dfrac{20}{30})^3=1-(0.67)^3=0.70$

Hence, the probability that he will get at least one lemon is 0.70.

5 0

3 years ago

The measured weights of 1,000 men in a certain village follow a normal distribution with a mean of 150 pounds and a standard dev

earnstyle [38]

Z = (X-Mean)/SD
z1 = (165 - 150)/15 = +1 
z2 = (135 - 150)/15 = - 1 
According to the Empirical Rule 68-95-99.7 
Mean +/- 1SD covers 68% of the values 
100% - 68% = 32% 
The remaining 32% is equally distributed below z = - 1 and z = +1 
32%/2 = 16%

Therefore,

a) Number of men weighing more than 165 pounds = 16% of 1000 = 160 
b) Number of men weighing less than 135 pounds = 16% of 1000 = 160

4 0

3 years ago

Read 2 more answers

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 $k\in\mathbb{R}$

(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)).

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4 years ago

7. A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use

Ulleksa [173]

Answer:

What company

Step-by-step explanation:

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3 years ago