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2 years ago

How to solve linear equations with variables on both sides of a triangle

Mathematics

1 answer:

sweet-ann [11.9K]2 years ago

7 0

Answer:

Ex of Linear Equation Of Both Sides

7x+8=3x+12

Step-by-step explanation:

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What is the area of the base of a triangular prism if the volume is 210 ft3 and the height is 14 ft?

Oksi-84 [34.3K]

We have the area of the base of a triangular prism= a*b
the volume of triangular prism =a*b*c( where c is the height) so that 210=a*b*14 ==> 210= base area*14==> base area= 210/14

3 0

3 years ago

Solve the inequality

GenaCL600 [577]

-1 < c + 2 < 3....subtract 2 from all sections
-1 - 2 < c + 2 - 2 < 3 - 2...simplify
-3 < c < 1
==============================
32 > 16 - 4g > 12....subtract 16 from all sections
32 - 16 > 16 - 16 - 4g > 12 - 16....simplify
16 > -4g > -4 ...now divide all sections by -4, and change inequality signs
16/-4 < (-4/-4)g < -4/-4...simplify
-4 < g < 1
============================
6y + 1 > = 10
6y > = 10 - 1
6y > = 9
y > = 9/6 which reduces to 3/2 or 1 1/2

-3/2y > = 9 ....multiply both sides by -2/3, cancelling the -3/2 on the left...and change the inequality sign
y < = 9 * -2/3
y < = -18/3 which reduces to - 6

so y > = 9/6(or 1 1/2) or y < = -6

3 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago

1. There are 78 sophomores at a school. Each is required to take at least one year of either chemistry or physics, but they may

SVETLANKA909090 [29]

We are given that there are a total of 78 students. If we set the following variables:

$\begin{gathered} C=\text{students only in chemestry} \\ P=\text{students only in physics} \\ PC=\text{students in physics and chemistry} \end{gathered}$

Then, the sum of all of these must be 78, that is: