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3 years ago

8

a student rolled two dice. what is the probability that the first die landed on a number less than 3 and the second die landed o

n a number greater than 4?

1 answer:

Oduvanchick [21]3 years ago

7 0

For the first die, the probability that the number is less than 3 is 2/6 or 1/3 because there are only 2 numbers (1 and 2) which are less than 3. For the second die, the probability that the number is greater than 4 is also 1/3 for only the numbers 5 and 6 satisfy the condition. To obtain the total probability, multiply the two probabilities because of the conjunction "and".

Thus, the answer is 1/9.

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Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) <u>Expand</u> <u>a</u> <u>polynomial</u> (or equation) is to multiply all the terms, remiving the parenthesis. <u>Reduce</u> <u>a</u> <u>polynomial</u> (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$

$S=10x^{2}-18x-15x+27$ (expand)

$S=10x^{2}-33x+27$ (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12+27$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x+15$

$20x^{2}-66x+27=0$

Solving:

$\frac{66+\sqrt{(-66)^{2}-(4.20.27)} }{2(20)}$

$\frac{66+46.86}{40}$

$x_{1}=\frac{66+46.86}{40}=$ 2.82

$x_{2}=\frac{66-46.86}{40}$ = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82

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