Which is the asymptote of the graph

iren2701 [21]

The correct answer to this question will be the yellow box which is <5 less than five

5 0

2 years ago

How has the school closing effected your daily life? How do you feel about learning math online? Be sure to mention some benefit

Zigmanuir [339]

Answer:

Step-by-step explanation:

While some might feel that online learning is the fun leisurely form of schooling that grants tons of free time, it honestly gets old quick. Being in quarantine makes it all the more worse because you do nothing all day. On the subject of math though, it is a disservice to have to learn math online. Personally, it is my experience that having a physical teacher and classroom to learn with is crucial to ones mathematical ability, but it's not the end of the world, it won't last long and it won't affect you that bad. Sure, the benefits include having more control over your day which is a good source of preparation for college life and responsibilities of being an adult, but you lose daily interpersonal interactions and typical casual daily exercise, since you no longer walk to class. Nevertheless, we will push on.

4 0

2 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

Parallelogram ABCD has the angle measures shown. Can you conclude that it is a rhombus, a rectangle, or a square? Explain.

Hoochie [10]

Answer:

The correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles

Step-by-step explanation:

In triangle ABD:

∠B = ∠D

Thus AB=AC by the property of opposite sides of equal angles are equal

In triangle CBD

∠B = ∠D

Thus CB=CD by the property of opposite sides of equal angles are equal

Thus all four sides of quadrilateral ABCD are equal

And diagonal BD bisects the angles

So, it is a rhombus

Therefore the correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles....

5 0

3 years ago

How do you estimate 52% of 84?

Nastasia [14]

The way to answer this question is to first make 52% into a decimal:

52% = 0.52

so now we will simply multiply:

84
x.52

once you have multiplied these numbers it comes out to be 43.68.
So your answer is 43.68.

Hope this answer helps! feel free to ask any additional questions :)

6 0

2 years ago