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Inessa05 [86]
3 years ago
10

Hiker's Elevation 3,000 Hiking The scatter plot shows a hiker's elevation above sea level during a hike from the base to the top

of a mountain. The equation of a trend line for the hiker's elevation is y=9.84x + 651, where x represents the number of minutes and y represents the hiker's elevation in feet. Use the equation of the trend line to estimate the hiker's elevation after 140 minutes. ​
Mathematics
1 answer:
iVinArrow [24]3 years ago
4 0

Answer:

2023

Step-by-step explanation:

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Sinθ+cotθ×cosθ=cscθ​
Assoli18 [71]

Answer:

Step-by-step explanation:

\sin \theta+\cot \theta\cos \theta=\sin \theta+\displaystyle \frac{\cos \theta}{\sin \theta}\cos \theta=\frac{\sin^2 \theta+\cos ^2 \theta}{\sin \theta}=\\\frac{1}{\sin \theta}=\csc \theta

5 0
2 years ago
Find the real or imaginary solutions of each equation by factoring
anyanavicka [17]
Sum of  2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so

x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4

the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
for x²-4x+16=0
x=\frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)}
x=\frac{4+/- \sqrt{16-64} }{2}
x=\frac{4+/- \sqrt{-48} }{2}
x=\frac{4+/- (\sqrt{-1})(\sqrt{48}) }{2}
x=\frac{4+/- (i)(4\sqrt{3}) }{2}
x=2+/- 2i\sqrt{3}


the roots are
x=-4 and 2+2i√3 and 2-2i√3
8 0
3 years ago
Ashley is preparing for a horse riding competition. She has trained her horse for 5 hours and has completed 150 rounds. At what
maxonik [38]
150/5 = 30

30 pounds per hour :)

I hope this helps!
3 0
2 years ago
Read 2 more answers
What percent of 200 is 290
Crazy boy [7]
145%
200/200: 100%
90/200= 45%
3 0
3 years ago
Read 2 more answers
Suppose a candidate for public office is favored by only 48% of the voters. if a sample survey randomly selects 2500 voters, the
Drupady [299]
Mean=0.48
standard deviation=0.01
thus using the z-score:
P(x>0.5) we shall have the following:
z=(0.5-0.48)/0.01=2
thus
P(x>0.5)
=1-P(x<0.5)
=1-P(z<2)
=1-0.9772
=0.0228
7 0
3 years ago
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