An advertisement says that 4 out of 5 doctors prefer a certain product. what percent of doctors do not like that product
2 answers:
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Answer:
Step-by-step explanation:
We will use slope-intercept form of equation to write our equation. The equation of a line in slope-intercept form is: , where m= Slope of the line, b= y-intercept.
To write the equation that represents the number of credits y on the cards after x games, we will find slope of our line.
We have been given that after playing 5 games we have 33 credits left. We play 4 more games and we have 21 credits left. So our points will be (5,33) and (9,21).
Let us substitute coordinates of our both given points in slope formula: ,
Now let us substitute m=-3 and coordinates of point (5,33) in slope intercept form of equation to find y-intercept.
Upon substituting m=-3 and b=48 in slope-intercept form of an equation we will get,
Therefore, our desired equation will be .
<h3>Answer: 24/25</h3>
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Explanation:
Sine is given to be negative, and so is tangent. This only happens in quadrant Q4
Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.
If tan(theta) < 0, then this means cos(theta) > 0
So we have y < 0 and x > 0 which places the angle somewhere in Q4.
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Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24
So this is what your steps may look like
a^2+b^2 = c^2
7^2+b^2 = 25^2
b^2+49 = 625
b^2 = 625-49
b^2 = 576
b = sqrt(576)
b = 24
So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25
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As an alternative, you could use the trig identity
sin^2(x) + cos^2(x) = 1
and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.
So,
sin^2(x) + cos^2(x) = 1
(-7/25)^2 + cos^2(x) = 1
(49/625) + cos^2(x) = 1
cos^2(x) = 1 - (49/625)
cos^2(x) = (625/625) - (49/625)
cos^2(x) = (625-49)/625
cos^2(x) = 576/625
cos(x) = sqrt(576/625)
cos(x) = sqrt(576)/sqrt(625)
cos(x) = 24/25
This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)