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3 years ago

I don’t know how to solve this

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

Answer:

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What does x equal???????? please help

Sholpan [36]

<h3>Answer: x = 61</h3>

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Explanation:

The angle x and the 29 degree angle combine to form a 90 degree angle. This is because the square maker on the left has that angle at 90 degrees, and all of the angles combine to form 180. So 180-90 = 90 is the left over amount.

Add up x and 29 to get 90

x+29 = 90

Solve for x by subtracting 29 from both sides

x+29-29 = 90-29

x+0 = 61

x = 61

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An alternative is to solve the equation below for x

x+29+90 = 180 ... see note below

x+119 = 180

x+119-119 = 180-119 ... subtract 119 from both sides

x = 61

we get the same answer

note: this equation turns into x+29 = 90 if you subtracted 90 from both sides

7 0

3 years ago

HELP FAST PLS I’ll mark u brainliest

Vesnalui [34]

Answer:

about 20 cents

Step-by-step explanation:

4 0

2 years ago

Which graph has a slope of 4/5 ?

Crank

Answer:

quadratic

Step-by-step explanation:

maybe you can try searching in google

8 0

2 years ago

Read 2 more answers

Divide 4/5 by 2/3.<br> a. 8/15b. 5/6c. 1 1/5d. 1 7/8

Ivan

4/5÷2/3=x
4/5×3/2=x
12/10=x
1 1/5=x

5 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