All categories

3 years ago

Find the distance between two points (2, -3) and (2,4).

Mathematics

1 answer:

Damm [24]3 years ago

6 0

Answer:

The distance between the two points is 7 units

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

(2, -3) and (2,4)

substitute the values in the formula

$d=\sqrt{(4+3)^{2}+(2-2)^{2}}$

$d=\sqrt{(7)^{2}+(0)^{2}}$

$d=7\ units$

You might be interested in

<img src="https://tex.z-dn.net/?f=%28fog%29%28%20-%205%29" id="TexFormula1" title="(fog)( - 5)" alt="(fog)( - 5)" align="absmidd

Xelga [282]

That should be the answer

8 0

3 years ago

Help me please

Nimfa-mama [501]

They are both in standard form.

7 0

3 years ago

Can someone say what 45-49 is with regrouping? thanks!

muminat

Answer:

it's -4 i don't know what you mean by regrouping, sorry

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Which shows one way to determine the factors of 4x3 + x2 – 8x – 2 by grouping?

lara31 [8.8K]

Group first 2 and group last 2 and find factors and undistribute

(4x^3+x^2)+(-8x-2)
(x^2)(4x+1)+(-2)(4x+1)
x^2(4x+1)-2(4x+1)
answer is first one

6 0

4 years ago

Read 2 more answers

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

4 years ago