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2 years ago

Which is the graph of y = |3x| – 2?

Mathematics

1 answer:

Rudik [331]2 years ago

6 0

Answer:

It's the 3rd one, that you've selected in the picture.

Step-by-step explanation:

Used a graphing calculator to check,

Hope this is helpful.

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There were 13426 eligible votes for a certain election.One day,8206 people cast their votes.What was the percent of voter turn o

coldgirl [10]

<h2>Greetings!</h2><h3>A way to do this is to divide the actual amount by the predicted amount:</h3>

8206 ÷ 13,426 = 0.6112

<h3>To change this back into a percentage simply multiply by 100:</h3>

0.6112 x 100 = 61.1

<h3>So the percent of voter turn out was 61.1%</h3><h2>Hope this helps!</h2>

4 0

3 years ago

What is 7x2/3 equal to

notka56 [123]

First you have to make 7 a fraction, every fraction with the denominator 1 is a whole number, so 7 should be 7/1,

7/1×2/3
=14/3
=4 2/3

answer=4 2/3
hope this helps....

8 0

3 years ago

Read 2 more answers

9/17 in decimal form rounded to the nearest hundredth

Elza [17]

Answer:

0.53

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

3,9,6,9,27,24,27,81,78,81,243,?find out last figure to complete series

Tatiana [17]

The next number is 240

6 0

3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago