Write the expanded form of 24.357 using fractions or decimals

PLEASE PLEASE PLEASE HELP! Please explain step by step!! Please please I'll give enough points!

ArbitrLikvidat [17]

1)
The distance between points A and B is;
$d= \sqrt{(x2-x1)^{2} + (y2-y1)^{2} } \\
d= \sqrt{(1-5)^{2} +(12+8)^{2}}\\ 
d= \sqrt{(-4)^{2} + (20)^{2}}\\
d=\sqrt{16+400}\\
d=\sqrt{416}\\
d=4\sqrt{26}$ ,
and if you wanted to you could measure out the legs of that hypotenuse using the slope between A and B, but I'm going to show you the easier way;
Δx = 4
Δy = -20
So from the midpoint, add 4 to the x-value and subract 20 from the y-value.
{(5+4),(-8-20)}
(9,-28) is the endpoint.

2)
Do the same thing as I did above;
$d= \sqrt{(x2-x1)^{2} + (y2-y1)^{2} } \\ 
d= \sqrt{(5-(-3))^{2} +(5-(-4))^{2}}\\
d=\sqrt{(5+8)^{2}+(5+4)^{2}}\\
d=\sqrt{(13)^{2}+(9)^{2}}\\
d=\sqrt{169+81}\\
d=\sqrt{250}\\
d=5\sqrt{10}$
So $5\sqrt{10}$ is the distance.

Sorry for the delay, life delayed me a ton ,:D

4 0

4 years ago

Math question help me

Rufina [12.5K]

I’m pretty sure it’s D

4 0

3 years ago

g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability th

xxTIMURxx [149]

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Sampling with replacement:

I consider a success choosing a black ball, so $p = \frac{50}{150+50} = \frac{50}{200} = 0.25$

We want 2 black balls and 2 white, 2 + 2 = 4, so $n = 4$ , and we want P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109$

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so $N = 200$

Sample of 4, so $n = 4$

50 are black, so $k = 50$

We want P(X = 2).

$P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116$

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

3 0

3 years ago

Simplify the surd 3√ 20 +√ 45

melisa1 [442]

Answer:

9 $\sqrt{5}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Simplify the radicals

$\sqrt{20}$

= $\sqrt{4(5)}$

= $\sqrt{4}$ × $\sqrt{5}$

= 2 $\sqrt{5}$

$\sqrt{45}$

= $\sqrt{9(5)}$

= $\sqrt{9}$ × $\sqrt{5}$

= 3 $\sqrt{5}$

Then

3 $\sqrt{20}$ + $\sqrt{45}$

= 3(2 $\sqrt{5}$ ) + 3 $\sqrt{5}$

= 6 $\sqrt{5}$ + 3 $\sqrt{5}$

= 9 $\sqrt{5}$

3 0

3 years ago

Read 2 more answers

Help yall, please! thanks

Setler79 [48]

For Problem #2, the second one should be 6x+4x+12, and the third one should be 10x+12

3 0

2 years ago