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3 years ago

How many solutions does the equation have?

Mathematics

2 answers:

Georgia [21]3 years ago

8 0

$3(4x-3)-7x=5x-9\\
12x-9-7x=5x-9\\
5x-9=5x-9\\
x\in \mathbb{R} \Rightarrow B$

Dovator [93]3 years ago

7 0

$3(4x-3)-7x=5x-9\\\\3(4x)+3(-3)-7x=5x-9\\\\12x-9-7x=5x-9\\\\\boxed{5x-9=5x-9}\\\\conclusion:\boxed{B.\ infinite\ solutions}$

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Orlov [11]

Answer:

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Step-by-step explanation:

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3 years ago

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yarga [219]

Answer:

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3 years ago

At a certain high school, 15 students play stringed instruments and

Aleks [24]

Answer:

0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N( $A \cup B$ )' = 5

To find:

The probability that a randomly selected students plays both = ?

Solution:

Total Number of students = N(A)+N(B)+N( $A \cup B$ )' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

$N(A\cap B) =0$

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$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0$

So, 0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

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