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victus00 [196]
3 years ago
15

The number of people attending the first basketball game of the season was 840. The number of people attending the last game of

the season was 1,200. What was the percent increase in attendance, to nearest percent?​
Mathematics
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

Step-by-step explanation:

percent increase = (new number - original number) / (original number) * 100

                             = (1200 - 840) / (840) * 100

                             = (360) / 840 * 100

                              = 0.4285 * 100

                              = 42.85% rounded to nearest percent = 43% <==

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What is the value of log2 18 rounded to the nearest hundredth?​
ZanzabumX [31]

Answer: 4,170

Step-by-step explanation:

\displaystyle\  \Large \boldsymbol{} log_2 18=log_2 \ 2+log_2 \ 9=1+3,16\underline99 \approax \approx4,170

6 0
2 years ago
I don't understand e. Please help
dexar [7]
Case 1: Probabilities cannot add up to a number greater (or less) than 1. This would mean there is greater than a 100% chance of something happening which just doesn't make sense. 0.4 + 0.4 + 0.3 = 1,1

Case 2: You cannot have a negative probability. That is claiming that there is a -10% chance of an event happening, there is at the very least a 0%. Despite them "adding" up to 1, the negative probability makes no sense.

Hope I helped!
6 0
2 years ago
The equation of the axis of symmetry is x = 3. True or false
Vsevolod [243]

Answer:

true

Step-by-step explanation:

6 0
3 years ago
Select all the correct answers.
MA_775_DIABLO [31]

Answer:

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4 0
3 years ago
Find the equation of the tangent line to the curve (a lemniscate)
olya-2409 [2.1K]

Answer:

m=\frac{9}{13} and b=\frac{40}{13}

Step-by-step explanation:

The equation of curve is

2(x^2+y^2)^2=25(x^2-y^2)

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})

4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})

The point of tangency is (-3,1). It means the slope of tangent is \frac{dy}{dx}_{(-3,1)}.

Substitute x=-3 and y=1 in the above equation.

4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})

40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})

-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}

80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240

130\frac{dy}{dx}=90

Divide both sides by 130.

\frac{dy}{dx}=\frac{9}{13}

If a line passes through a points (x_1,y_1) with slope m, then the point slope form of the line is

y-y_1=m(x-x_1)

The slope of tangent line is \frac{9}{13} and it passes through the point (-3,1). So, the equation of tangent is

y-1=\frac{9}{13}(x-(-3))

y-1=\frac{9}{13}(x)+\frac{27}{13}

Add 1 on both sides.

y=\frac{9}{13}(x)+\frac{27}{13}+1

y=\frac{9}{13}(x)+\frac{40}{13}

Therefore, m=\frac{9}{13} and b=\frac{40}{13}.

5 0
2 years ago
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