(3x^3 - 5x^2 -4x + 4) / (x-2) = (3x-2)(x+1) when x ≠2
Josephine travels 32500+78000 yards at 60 mph
=110500yards
=62.78miles
time = distance/spped
time = distance/60
time=62.78/60
time=1.05 (1 hour 5 mins)
Marcus travels 78000(2)(square) +32500(2)
= 84500yards
=48 miles
time = distance/speed
time=48/45
time=1.07 (1 hour 7 mins)
a)Josephine arrives first
b)she arrives earlier by 2 mins.
Answer:
- the domain is [0,129]
- 112 mm
- 1942
Step-by-step explanation:
1. The function is good for years 1880 to 2009, 0 to 129 years after 1880. Values of x can be anything in the domain [0, 129].
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2. 1950 -1880 = 70. The year 1950 corresponds to x=70, so the function tells us the water level rose ...
f(70) = 1.6·70 = 112 . . . . . mm
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3. We want to find x when f(x) = 100. That will be the solution to ...
100 = 1.6x
100/1.6 = x = 62.5
Then 62.5 years after 1880 is year 1942.5. The water level was 100 mm higher than in 1880 in the year 1942.
(1)
Mean length of all fish in the sample - 
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 3 + 5 + 8 + 2 + 10 + 9 + 4 + 4)/(10)
= 62/10
=<em> 6.2</em>
(2)
Mean length of adult fish in the sample - 
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (12 + 5 + 8 + 10)/(4)
= 35/4
=<em> 8.75</em>
(3)
Mean length of juvenile fish in the sample - 
<u>Mean</u> = (Sum of observations)/(Number of observations)
= (5 + 3 + 2 + 9 + 4 + 4)/(6)
= 27/6
<em>= 4.5</em>
(4)
Percentage of sample that were adult fish - 
<u>Percentage</u> = (No. of adult fishes)/(Total no. of fishes) × 100
% = (4/10) × 100
<em>% = 40</em>
(5)
Percentage of sample that were juvenile fish - 
<u>Percentage</u> = (No. of juvenile fishes)/(Total no. of fishes) × 100
% = (6/10) × 100
<em>% = </em><em>6</em><em>0</em>
(6)
Percentage of sample that were juveniles over 8 inches long - 
<u>Percentage</u> = (No. of juveniles over 8 inches)/(Total no. of fishes) × 100
% = (1/10) × 100
<em>% = </em><em>1</em><em>0</em>
Answer:
so they dont get cramps or chairly horses
Step-by-step explanation: