All categories

2 years ago

Make x the subject of the formula х=y - 3 4z

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

7 0

Answer:

it's already been subjected

This is core maths or elective maths?

You might be interested in

Can anyone help me with this plz

inn [45]

Answer:

Step-by-step explanation:

So we know right off the bat the angles ABE and CBD are congruent because of vertical angles.

Now we can look at the sides.

The ratio for 27:18 = 3:2

The ratio for 36:24 = 3:2

They check out.

That means that the answer is B. SAS

5 0

3 years ago

What equation relates the number of quarters x and the number of quarters x and the number of dime y to the goal of $10

Dimas [21]

.25x+.10y=$10

.25 because that is the value of the quarter, times x the number or quarters you have and .10 because that is the value of the dime times y, the number of dimes you have.

8 0

3 years ago

Huixian needs to pack 171 pens, 63 pencil, and 27 erasers into identical bags so that each item is equally distributed among the

luda_lava [24]

Answer:

87 gift bags

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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 absolute deviation of 4 in this data set?

lakkis [162]

Answer:

Step-by-step explanation:

I took the k12 quiz

6 0

2 years ago