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alexgriva [62]
3 years ago
11

8 percent of what number is 2

Mathematics
1 answer:
Sedbober [7]3 years ago
5 0
8%×n=2. 2÷8/100=2×100/8=100/4=25. so the answer is 25.
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Tell whether the points appear to represent a linear function, an exponential function, or neither? Explain please.
mario62 [17]

Answer:

Exponential

Step-by-step explanation:

Linear is easy to remember because it goes in a straight LINE.

Only other option is a quadratic and this does not create a parabola

8 0
2 years ago
$87.00 divided by 5 can u divide pls
Illusion [34]
$87.00/5 is $17.4 dollars
6 0
3 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
Solve for x.<br> <br> 13x + 7 = 5x - 20
myrzilka [38]
The answer is -27/8= -3.375
7 0
3 years ago
Read 2 more answers
The times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a standard deviation
serious [3.7K]

The probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

<h3>What is normally distributed data?</h3>

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data.

The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The times of the runners in a marathon are normally distributed, with

  • Mean of 3 hours and 50 minutes
  • Standard deviation of 30 minutes.

Refere the probabiliity table attached below. The probability of Z being inside the 1 Standard daviation of mean is 0.84.

The probability of runner selected with time less than or equal to 3 hours and 20 minutes,

P=1-0.84\\P=0.16

Thus, the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

Learn more about the normally distributed data here;

brainly.com/question/6587992

4 0
2 years ago
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