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marissa [1.9K]
3 years ago
12

Which of the following situations represent linear growth? Check all of the boxes that apply.

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
8 0

Answer:

A and D

Step-by-step explanation:

3% each year changes. 3% for 100 is not the same as 3% for 103. Losing weight is a decrease, not growth.

garri49 [273]3 years ago
8 0

Answer:

A, C, and D

Step-by-step explanation:

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4x - 2 (x-2) = -4 + 5x -13
kolezko [41]

Answer:

x = 7

Step-by-step explanation:

Given

4x - 2(x - 2) = - 4 + 5x - 13 ← distribute left side and simplify

4x - 2x + 4 = 5x - 17

2x + 4 = 5x - 17 ( subtract 5x from both sides )

- 3x + 4 = - 17 ( subtract 4 from both sides )

- 3x = - 21 ( divide both sides by - 3 )

x = 7

8 0
3 years ago
From the measurements of the legs of a right triangle measuring 3cm, 4cm and the hypotenuse of the same 5cm, calculate the measu
WARRIOR [948]

Answer:

3cm and 4cm addition

7cm and 5cm subtract

2cm answer

3 0
3 years ago
The following lists the joint probabilities associated with smoking and lung disease among 60-to-65 year-old men. Has Lung Disea
Anna71 [15]

Answer:

4.11% probability that he has lung disease given that he does not smoke

Step-by-step explanation:

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Does not smoke

Event B: Lung disease

Lung Disease/Nonsmoker 0.03

This means that P(A \cap B) = 0.03

Lung Disease/Nonsmoker 0.03

No Lung Disease/Nonsmoker 0.7

This means that P(A) = 0.03 + 0.7 = 0.73

What is the probability of the following event: He has lung disease given that he does not smoke?

P(B|A) = \frac{0.03}{0.73} = 0.0411

4.11% probability that he has lung disease given that he does not smoke

5 0
3 years ago
Somebody please help with this problem
Mrrafil [7]

Step-by-step explanation:

We have been given that AE=BE and \angle1\cong \angle2.

We can see that angle CEA is vertical angle of angle DEB, therefore, m\angle CEA=m\angle DEB as vertical angles are congruent.

We can see in triangles CEA and DEF that two angles and included sides are congruent.

\angle 1\cong \angle 2

AE=BE

\angle CEA\cong\angle DEB or \angle 3\cong \angle 4  

Therefore, \Delta CEA\cong \Delta DEB by ASA postulate.

Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.

5 0
3 years ago
(2t^2+13t+15)\(t+6) long division
aleksklad [387]

Answer:

I think this is the correct solution

6 0
3 years ago
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