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3 years ago

I need this done it’s due tmr and I need each problem simplified in the lowest fraction

Mathematics

1 answer:

Ivenika [448]3 years ago

8 0

Answer:

photomath

Step-by-step explanation:

theres an app where you can take a pic of your problem and it would solve it for you,including fractions. I'm pretty sure it will simplify your answers too.The app is called photomath

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What is x, the distance between points A and A'? 2. 4 units 4. 8 units 13. 6 units 14. 4 units.

adelina 88 [10]

The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

Given that,

The points A (3, -5) and A' (2, -3).

We have to determine,

The distance between point A and A'.

According to the question,

The distance between two points is determined by using the distance formula.

$\rm Distance \ formula = \sqrt{(x_2-x_2) ^2+ (y_2-y_2)^2$

Then,

The distance between points A (3, -5) and A' (2, -3) is,

$\rm Distance = \sqrt{(2-3) ^2+ ((-3)-(-5))^2}\\\\ Distance = \sqrt{(-1) ^2+ (2)^2}\\\\ Distance = \sqrt{1+4}\\\\Distance = \sqrt{5}\\\\Distance = 2.4 \ units$

Hence, The distance between points A (3, -5) and A' (2, -3) is 2.4 units.

For more details refer to the link given below.

brainly.com/question/8069952

6 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

HELP I NEED HELP ASAP

Nady [450]

Answer:

Step-by-step explanation:

because the -2 is where the point is and the slope will start there so when you draw a line the line will pass by (0,-2)

5 0

2 years ago

End of year assessment

tensa zangetsu [6.8K]

Answer:

Don't fail

Step-by-step explanation:

4 0

2 years ago

And here is the other one.

LUCKY_DIMON [66]

Since BD bisects angle ABC, that means angle ABD and angle CBD are equal to each other. With that set up the equation to solve for x like this:

-4x+33 = 2x+81
-2x -2x
————————
-6x +33 = 81
-33 -33
————————
-6x = 48
————- (divide by -6)
-6

x = -8

Now substitute that to ABD
-4(-8) +33
32 +33
=65

here’s CBD
2(-8) + 81
-16 + 81
=65

Finally angle ABC will be double the amount of ABD or CBD so 65 times 2 is 130.

ANSWERS: (angles)
ABD and CBD: 65
ABC: 130

3 0

3 years ago