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3 years ago

what is the portion of a line that starts at one point and goes off in a particular direction to infinity

Mathematics

2 answers:

krok68 [10]3 years ago

6 0

a line segment neither starts nor ends at ant certain point and doesnt have arrows at either ends

snow_lady [41]3 years ago

3 0

Answer:

C. Ray

Step-by-step explanation:

Ray because it is the portion of a line that starts at one point and goes off in a particular direction to infinity?

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Anyone help? I'm trying to answer this question but I'm confused.

xeze [42]

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Bbhhbjjbvfgbbyy Jill

3 0

3 years ago

The vertex of a function is (-3, 0) and is a minimum. What is the axis of symmetry and how is it determined?

nalin [4]

The axis of symmetry would be -3 because it is always the (x) of the vertex

5 0

2 years ago

Ms Melissa and Ms ryanne went for a walk and they got back at 8:02am their walk was actually 45 minutes what time did they leave

dimaraw [331]

Answer: 7: 17

Step-by-step explanation:

This doesn't really make sense but it's how I do it

8:02 = 7:62

7:62

7: 17

6 0

3 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

What is the area of this figure?

Paraphin [41]

Area of square
= 7x7
=49
Area of triangle
=0.5x 7x12
=42
Total area of figure
=49+42
=91

Hence, answer is D

6 0

3 years ago