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4 years ago

What is the approximate distance between the points (-3, -4) and (-8, 1) on a coordinate grid

Mathematics

1 answer:

motikmotik4 years ago

7 0

Answer:

$D= 7.071$

Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

Here we are going to use the distance formula, which is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (-3,-4))$

$(x_2,y_2) = (-8,1)$

Replacing these values in the distance formula

$D =\sqrt{(-8-(-3))^2+(1-(-4))^2}$

$D =\sqrt{(-8+3)^2+(1+4)^2}$

$D =\sqrt{(-5)^2+(5)^2}$

$D =\sqrt{25+25}$

$D =\sqrt{50}$

$D= 7.071$

Hence the Distance is 7.071 units

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3 years ago

-3 squared times 4 divided by 2 over square root of 9-2 square, and then the sum multiplied by the cube root of 25

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So you want to remember PEMDAS.
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3 years ago

Anna has fabric that is 1/4 yard long. She needs to cut the fabric into pieces that are 1/8 yard long. How many 1/8 yard long pi

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3 years ago

There are three urns that each contain 10 balls. The first contains 3 white and 7 red balls, the second 8 white and 2 red, and t

malfutka [58]

Answer:

(a) 0.0520

(b) 0.3699

Step-by-step explanation:

Let $E_{1},E_{2},E_{3}$ be the events of selecting urn 1, urn 2 and urn 3 respectively.

So, $P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}$

Let W be the event of drawing a white ball.

Now, probability of drawing 2 white balls from urn 1 is given as:

$P(W\cap W \cap E_{1})=P(E_{1})\times P((W\cap W)/E_{1})\\P(W\cap W \cap E_{1})=\frac{1}{3}\times (\frac{3}{10})^{2}=\frac{9}{300}$

Probability of drawing 2 white balls from urn 2 is given as:

$P(W\cap W \cap E_{2})=P(E_{2})\times P((W\cap W)/E_{2})\\P(W\cap W \cap E_{2})=\frac{1}{3}\times (\frac{8}{10})^{2}=\frac{64}{300}$

Probability of drawing 2 white balls from urn 3 is given as:

$P(W\cap W \cap E_{3})=P(E_{2})\times P((W\cap W)/E_{2})=\frac{1}{3}\times 1=\frac{1}{3}$ ............. (as all balls are white only)

Now, probability of drawing 2 white balls is the sum of all the above probabilities and is given as:

$P(2W)=P(W\cap W \cap E_{1})+P(W\cap W \cap E_{2})+P(W\cap W \cap E_{3})\\P(2W)=\frac{9}{300}+\frac{64}{300}+\frac{1}{3}\\P(2W)=\frac{9}{300}+\frac{64}{300}+\frac{100}{300}=\frac{9+64+100}{300}=\frac{173}{300}$

(a)

Probability of selecting urn 1 given that 2 white balls are drawn is:

$P(E_{1}/2W)=\frac{P(E_{1})\times P(2W/E_{1})}{P(2W)}\\P(E_{1}/2W)=\frac{\frac{9}{300}}{\frac{173}{300}}=\frac{9}{173}=0.0520$

Therefore, probability of selecting urn 1 given that 2 white balls are drawn is 0.0250.

(b)

Probability of selecting urn 2 given that 2 white balls are drawn is:

$P(E_{2}/2W)=\frac{P(E_{2})\times P(2W/E_{2})}{P(2W)}\\P(E_{2}/2W)=\frac{\frac{64}{300}}{\frac{173}{300}}=\frac{64}{173}=0.3699$

Therefore, probability of selecting urn 2 given that 2 white balls are drawn is 0.3699

8 0

4 years ago