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anyanavicka [17]
3 years ago
11

There are 13 students in a biology class and 12 students in a chemistry class. if there are 20 students in total, how many stude

nts are in both classes?
Mathematics
1 answer:
klemol [59]3 years ago
4 0

There are 5 students in both classes. Have a great day!

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Grant thinks he has reached the point where he has maximized his profit. However, because this can be tricky to determine, he is
Mashcka [7]

Answer:c

Step-by-step explanation:

Profit maximization happens with marginal revenue is equal to marginal cost, so if Grant's assumption was right before selling the extra unit, when he actually sells the extra unit, this will increase his revenue

That's why the answer is c

Marginal revenue will exceed marginal cost

6 0
3 years ago
16 + 2q = 6 what’s this?
nika2105 [10]

Answer:

6 - 16 = -10

-10/2 = -5

q = -5

4 0
3 years ago
Graph the following inequality. Then click to show the correct graph.<br><br><br>3x - 2y ≤ 6
max2010maxim [7]

Answer:

We want to graph the inequality:

3x - 2y ≤ 6

The first step is to write this as a linear equation, to do it, we can isolate y in one side of the inequality.

3x  ≤ 6 + 2y

3x - 6  ≤ 2y

(3/2)x  - 6/2  ≤ y

(3/2)x  - 3  ≤ y

or:

y ≥ (3/2)x  - 3

Because we have the symbol ≥

The points on the line are solutions, then the first part is to graph the line:

y = (3/2)*x - 3

Next, we have:

y equal to or larger than  (3/2)*x - 3

Then we need to shade all the region above that line.

The graph can be seen below.

3 0
3 years ago
What type of correlation is shown in the graph ? positive no correlation linear negative
mario62 [17]

Answer:

positive

Step-by-step explanation:

8 0
3 years ago
Given that sin theta = 1/4, 0→theta→π/2, what<br>. what is the exact value of cos 8?​
natali 33 [55]
<h3>Answer: Choice B \frac{\sqrt{15}}{4}</h3>

===========================================================

Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

Square both sides of the given equation

\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}

Then use the pythagorean trig identity to get

\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\

3 0
3 years ago
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