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3 years ago

For the following represent the sample space in set notation and sate how many outcomes are

Mathematics

1 answer:

Sphinxa [80]3 years ago

3 0

Answer:

I got blue 3 and green 7

Step-by-step explanation:

why it is blue 3 is because there are 3 of the blue and for the green there are 4 it landed on the 3 green and that was the 7th one

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The triangle is created by the path that Julie walks from home, to school, to the park then back to home . How many miles is the

vladimir2022 [97]

To find a perimeter of something you just add all the outside measurements together
Which is:
2.4+2.1+1.5= 6 miles

3 0

3 years ago

You are going to use an incline plane to lift a heavy object to the top of a shelving unit with a height of 5 ft. The base of th

loris [4]

Answer:

15.8 ft

Step-by-step explanation:

The inclined plane, the base of and the shelving unit form the shape of a right angled triangle.

The hypotenuse is the length of the inclined plane, h.

The base of the triangle is 15 ft.

The height of the triangle is 5 ft.

To find the hypotenuse, h, we have to use Pythagoras rule:

$h^2 = a^2 + b^2$

where a = height of the triangle

b = base of the triangle

Therefore:

$h^2 = 5^2 + 15^2\\\\h^2 = 25 + 225 = 250\\\\h = \sqrt{250}\\ \\h = 15.8 ft$

The inclined plane is 15.8 ft long.

5 0

3 years ago

There are 10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles in a box. If two marbles are drawn at random with

Nikitich [7]

Answer:

7/20

Step-by-step explanation:

10 blue marbles, 4 black marbles, 5 white marbles, and 6 red marbles = 25 marbles

25-10 = 15

There are 15 not blue marbles

P( not blue ) = not blue marbles / total = 15/25 = 3/5

keep the not blue marble

25-1 = 24 marbles left

15-1 = 14 not blue marbles

P( not blue ) = not blue marbles / total = 14/24 = 7/12

P( not blue, keep, not blue) = 3/5 * 7/12 = 21/60 = 7/20

6 0

4 years ago

Read 2 more answers

Uestion

Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}$

now, let's check the other path, JM and KL

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}$

so the red path will be $5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare$

3 0

2 years ago

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago