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Alexus [3.1K]
3 years ago
15

Select all possible values for x if −3x < 15.

Mathematics
2 answers:
makkiz [27]3 years ago
4 0

Step-by-step explanation:

When dividing or multiplying by a negative number, remember to flip the inequality sign.

-3x < 15

x > -5 (divide -3 on both sides, flip the inequality sign)

The correct options are 2 and 9. (D and E)

Maru [420]3 years ago
4 0

Answer:

D. 2

E. 9

Step-by-step explanation:

- 3x < 15 \\  \\ x >  \frac{15}{ - 3} \\  \\ x  >  - 5 \\  \\

Since x > - 5,

Therefore possible values of x would be 2 and 9 out of the given ones.

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The municipality of Waterloo needs $915,000 from property tax to meet its budget. The total value of assessed property in Waterl
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Need help on probability. Please explain how you did it so I can know how to do it.
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Answer:

\text{P}(B \text{ or }C)=\dfrac{5}{6}

Step-by-step explanation:

<u>Mutually Exclusive Events</u>

For two events, A and B, where A and B are mutually exclusive:

\boxed{\text{P}(A \text{ or }B)=\text{P}(A)+\text{P}(B)}

<u>Probability distribution table</u>:

\begin{array}{|c|c|c|c|c|c|c|c|}\cline{1-7} x & 1 & 2 & 3 & 4 & 5 & 6 \\\cline{1-7} \text{P}(X=x)\phantom{\dfrac11}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\\\cline{1-7}\end{array}

where X is the score on a fair, six-sided dice.

P(B or C) means "the probability of rolling an even number or a multiple of 3".

As an even number of a fair, six-sided dice can never be a multiple of 3, the two events B and C are mutually exclusive.

Calculate the probabilities for events B and C.

<u>Event B</u>

Rolling an even number.

\begin{aligned}\implies \text{P}(X \text{ is even})&=\text{P}(2 \text{ or }4\text{ or }6)\\\\&=\text{P}(X=2)+\text{P}(X=4)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{3}{6}\end{aligned}

<u>Event C</u>

Rolling a multiple of 3.

\begin{aligned}\implies \text{P}(X \text{ is multiple of 3})&=\text{P}(3 \text{ or }6)\\\\&=\text{P}(X=3)+\text{P}(X=6)\\\\& = \dfrac{1}{6}+ \dfrac{1}{6}\\\\&=\dfrac{2}{6}\end{aligned}

<u>Solution</u>

\begin{aligned}\implies \text{P}(B \text{ or }C)&=\text{P}(B)+\text{P}(C)\\\\& = \dfrac{3}{6}+\dfrac{2}{6}\\\\& = \dfrac{5}{6}\end{aligned}

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