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Flauer [41]
3 years ago
12

4x-1=y graphed by using a table

Mathematics
1 answer:
olasank [31]3 years ago
5 0

Answer:

It can be many answers since it’s on a graph

Step-by-step explanation:

That equation isn’t the rule so test it out on number so and you will get point to put on a graph

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What is the range of this function?
Lostsunrise [7]

Answer:

c \: is \: correct \:  \:  \: the \: range \: is \: usually \: right \ \\ : and \: they \: are \: x \:

5 0
3 years ago
Which relationship is always correct for the angles x, y, and z of triangle ABC?
Zarrin [17]
Your answer is B. y+z=x.
7 0
3 years ago
Read 2 more answers
A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
2 years ago
1. 2(H-8)-h=h-16
podryga [215]

(1) C and (2) D

(1)

distribute the left side of the equation

2h - 16 - h = h - 16 ( simplify left side )

h - 16 = h - 16

Since both sides are equal, any value of h will make the equation true.

Hence there are infinitely many solutions to the equation → C

(2)

3 + 6z = 13 + 6z ( subtract 6z from both sides )

3 = 13 ← not possible

Hence there are no solutions to the equation → D



8 0
3 years ago
SOLVE: 6^3x+2 = 4^x-5
Lilit [14]
There's an app called photomath that would be useful for these kinds of problems!
6 0
3 years ago
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