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3 years ago

Can someone help me on 6 please

Mathematics

1 answer:

scZoUnD [109]3 years ago

7 0

Answer:

1) Inequality: 110x>150+85x

2) Solution: x>6, then Nadia would need to sell more than 6 ads per week in order for Choice A to be the better choice,

Solution:

Be x the number of ads Nadia sells

A) Choice A: $110 for each ad she sells. She would earn:

Ea=110x

B) Choice B: A weekly salary of $150, plus $85 for each ad she sells. She would earn:

Eb=150+85x

1) Write an inequality to determine the number of ads Nadia would need to seel per week in order for Choice A to be the better choice:

Ea>Eb

110x>150+85x

2) Solution

Solving for x: Subtracting 85x both sides of the equation:

110x-85x>150+85x-85x

Subtracting:

25x>150

Dividing both sides of the equation by 25:

25x/25>150/25

Dividing:

x>6

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Probability that a student chosen randomly from the class plays basketball or baseball is $\frac{23}{30}$ or 0.76

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Number of students who plays base both the games = 8

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Probability that a student chosen randomly from the class plays basketball or baseball=?

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$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ ---------------(1)

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P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) = $\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}$

P(A) = $\frac{19}{30}$ --------------------------(2)

Finding P(B)

P(B) = $\frac{\text { Number of students playing baseball }}{\text{Total number of students}}$

P(B) = $\frac{12}{30}$ ---------------------------(3)

Finding $P(A \cap B)$

P(A) = $\frac{\text { Number of students playing both games }}{\text{Total number of students}}$

P(A) = $\frac{8}{30}$ -----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

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$P(A \cup B)= \frac{23}{30}$

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