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CaHeK987 [17]
3 years ago
5

What is the probability that the first marble is blue and the second marble is yellow?

Mathematics
1 answer:
Cloud [144]3 years ago
5 0

Answer:

\frac{1}{14}

Step-by-step explanation:

<u><em>First time drawing:</em></u>

Total outcome is 8

Favorable outcome is 2

The probability that the first marble drawing in random is blue is

\frac{2}{8} = \frac{1}{4}

<u><em>Second time drawing (with no replacement): </em></u>

Total outcome is 7  (8 - 1 = 7)  

Favorable outcome is 2

The probability that the second marble drawing in random is yellow is

\frac{2}{7}

<em>The probability that the first marble is blue and the second marble is yellow is</em> \frac{1}{4} × \frac{2}{7} = \frac{2}{28} = \frac{1}{14}

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4. Option C (3a+2) inches

5. Extraneous solution x=-5/6 because we get for width and length negative values.

Solution: Value of x is 3

Length of the box: 5 ft

Width of the box: 4 ft

Step-by-step explanation:

4. Area of a rectangle: A=12a^2-a-6 square inches

Width: w=4a-3

Length: l=?

A=w l

Replacing A by 12a^2-a-6 and w by 4a-3

12a^2-a-6 = (4a-3) l

Solving for l: Dividing both sides of the equation by 4a-3:

(12a^2-a-6) / (4a-3) = (4a-3) l / (4a-3)

Simplifying:

(12a^2-a-6) / (4a-3) = l

l = (12a^2-a-6) / (4a-3)

Factoring the numerator:

12a^2-a-6 = (4a-3)(3a+2)

Let's check it:

(4a-3)(3a+2)=4a(3a)+4a(2)-3(3a)-3(2)=12a^2+8a-9a-6→(4a-3)(3a+2)=12a^2-a-6

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Simplifying:

l = (3a+2) inches


5. Length: l=(3x-5) ft

Width: w=(2x-1) ft

Height: h=2 ft

Volumen of the box: V=40 ft^3

x=?

Length: l=?

Width: w=?

V = l w h

Replacing the given:

40 ft^3 = (3x-5) ft (2x-1) ft 2 ft

40 ft^3 = 2 (3x-5)(2x-1) ft^3

40=2(3x-5)(2x-1)

Dividing both sides of the equation by 2:

40/2=2(3x-5)(2x-1)/2

Simplifying:

20=(3x-5)(2x-1)

Eliminating the parentheses on the right side of the equation applying the distributive property:

20=3x(2x)+3x(-1)-5(2x)-5(-1)

20=6x^2-3x-10x+5

Adding like terms:

20=6x^2-13x+5

Equaling to zero: Subtracting 20 from both sides of the equation:

20-20=6x^2-13x+5-20

0=6x^2-13x-15

6x^2-13x-15=0

ax^2+bx+c=0; a=6, b=-13, c=-15

Using the quadratic formula:

x=[-b+-sqrt(b^2-4ac)]/(2a)

x=[-(-13)+-sqrt((-13)^2-4(6)(-15))]/(2(6))

x=[13+-sqrt(169+360)]/12

x=[13+-sqrt(529)]/12

x=[13+-23]/12

x1=(13-23)/12=(-10)/12=-10/12=-(10/2)/(12/2)→x1=-5/6

x2=(13+23)/12=36/12→x2=3


With x=-5/6

l=(3x-5) ft

l=(3(-5/6)-5) ft

l=(-5/2-5) ft

l=-(5/2+5) ft

l=-(5+2(5))/2 ft

l=-(5+10)/2 ft

l=-15/2 ft < 0. The length cannot be a negative number then x=-5/6 is a extraneous solution.

w=(2x-1) ft

w=(2(-5/6)-1) ft

w=(-5/3-1) ft

w=-(5/3+1) ft

w=-(5+3(1))/3 ft

w=-(5+3)/3 ft

w=-8/3 ft <0. The width cannot be a negative number then x=-5/6 is a extraneous solution.


With x=3

l=(3x-5) ft

l=(3(3)-5) ft

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l=4 ft

w=(2x-1) ft

w=(2(3)-1) ft

w=(6-1) ft

w=5 ft

and h=2 ft

Let's check the volume

V= w l h

V=(5 ft)(4 ft)(2 ft)

V=40 ft^3 Correct

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