Hamburger costs $1.49 for 1/2 pound. How much would the hamburger cost for a whole pound?

A survey of pet owners shows that:

Tpy6a [65]

D have neither a cat nor a dog

8 0

2 years ago

Read 2 more answers

Which graph represents a reflection of f(x) = 6(0.5)x across the x-axis?

Musya8 [376]

The required graph is attached.

7 0

2 years ago

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The length of a rectangle Is 3 Inches more than its width, and its perimeter is 22 Inches. Find the width of the rectangle.

Vesnalui [34]

Answer:

the width of the rectangle is 7 inches

Step-by-step explanation:

3 0

2 years ago

Consider a circle that has a circumference of 28/3.14 centimeters. What is the area of this circle? Answer fast plz also can you

bija089 [108]

Answer:

$A=6.33\ cm^2$

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The circumference of a circle is equal to

$C=2\pi r$

we have

$C=\frac{28}{3.14}\ cm$

$\pi=3.14$

substitute the values

$\frac{28}{3.14}=2(3.14)r$

$r=\frac{28}{19.7192}\ cm$

$r=1.42\ cm$

step 2

Find the area of the circle

The area of the circle is equal to

$A=\pi r^{2}$

substitute the values

$A=(3.14)(1.42)^{2}$

$A=6.33\ cm^2$

7 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