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Igoryamba
3 years ago
7

A tree that is 2 feet tall is growing at a rate of 1 foot per year. A 4-foot tall tree is growing at a rate of 0.5 foot per year

. In how many years will the trees be the same height
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer:

4 years

Step-by-step explanation:

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3 years ago
Read 2 more answers
Select ALL the correct answers.
jolli1 [7]

Answer:

(4x+5)(-3x-1) = -12x²-19x-5

Option A:

   (-16x² + 10x - 3) + (4x² - 29x - 2) = -12x²-19x-5

   Option A is correct.

Option B:

   3(x - 5) - 2(6x² + 9x + 5) = -12x²-15x-25

   Option B is wrong.

Option C:

   2(x - 1) - 3(4x² + 7x + 1) = -12x²-19x-5

   Option C is correct.

Option D:

   (2x² - 11x - 9) - (14x² + 8x - 4) = -12x²-19x-5

   Option D is correct.

4 0
3 years ago
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
3 years ago
When should you use dot plots and why?
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3 0
2 years ago
Morgan is walking her dog on an 8-meter-long leash. She is currently 500 meters from her house, so the maximum and minimum dista
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Given:
Morgan = 500 meters from her house
Length of leash = 8 meters

minimum distance of the dog from the house 500 - 8 = 492 meters.
maximum distance of the dog from the house 500 + 8 = 508 meters.
7 0
3 years ago
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