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alexgriva [62]
3 years ago
7

Create or find your own, unique function h(x). discuss the limit of your function as latex: x x approaches some particular value

, let's say a, where the limit exists but the y-value does not or is different than the limit.
Mathematics
1 answer:
Zina [86]3 years ago
6 0
Consider the function:

h(x)= \frac{x^2-1}{x-1}

As x tends to 1, 

h(1)= \frac{1^2-1}{1-1} = \frac{0}{0}

hence, the y value does not exist.

But

\lim_{x\to1} h(x)= \lim_{x\to1}  \frac{x^2-1}{x-1}  \\  \\ = \lim_{x\to1} \frac{(x-1)(x+1)}{x-1}= \lim_{x\to1} (x+1) \\  \\ =1+1=2

Therefore, the limit of h(x) exists but the y-value does not exist.
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Flura [38]

Answer:

c

Step-by-step explanation:

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You are going to an amusement park over Spring Break.
agasfer [191]

Answer:

you can go on a maximum of 11 rides

Step-by-step explanation:

you first have to subtract 30 from 100 since that is the admissons fee.

So now you are left with 70 dollars.

You divide 70 by 6 since you need 6 dollars to go on each ride.

and 70 divided by 6 is 11.6666666667 but round it to 11 since that the whole number you're left with.

hope this helped :)

5 0
3 years ago
The surface area of a cube is 400 millimeters what is the surface area of a similar cube that is larger by a scale factor of 3
yuradex [85]
Larger by 3 means that the legnths of the sides are larger by 3 times
so we have a cube that has area of 400
surface area=legnth times width times 6 since it is a cube and legnth=width
so
SA=w^2 times 6

6w^2=400
divide by 6
w^2=200/3
square root both sides
w=√(200/3)
so w is 3 times larger
(3w)^2 times 6 is new area

w=√(200/3)
(3√(200/3))^2 times 6=9(200/3) times 6=600 times 6=3600

answer is 3600 ml^2
3 0
3 years ago
X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
1 year ago
4 Math Questions<br> 1.) a^2-10a=4<br> 2.) x^2+20x-18=3<br> 3.) (x-5)^2=1<br> 4.) x^2-18x=65
natima [27]

1) a=5+√29 or a=5−√29 2) x=1 or x=−21 3) x=4 or x=6 4) x=9+√146 or x=9−√146 Hopefully that helps you ❤

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3 years ago
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