We will be using this equation for this problem
d = ut + ½.at²
<span>Given:</span>
<span>initial velocity, u = 0 (falling from rest) </span>
<span>acceleration, a = +9.80 m/s²(taking down as the convenient positive direction) </span>
<span>Time = 1.0s, 2.0s, 3.0s, 4.0s, 5.0s </span>
<span>Using .. d = ½.at² each time (each calculation is the distance from the top) </span>
<span>For 1.0s .. d = ½ x 9.80 x (1²) = 4.90 m </span>
<span>For 2.0s .. d = ½ x 9.80 x (2²) = 19.60 m </span>
<span>3.0s .. d = 44.10m (you show the working for the rest) </span>
<span>4.0s .. d = 78.40 m </span>
<span>5.0s .. d = 122.50m </span>
<span>Plot distance (displacement from the top) on the y-axis against time on the x-axis (label axes and give units for each).The line of best fit will be a smoothly upward curving line getting progressively steeper. Do not join graph points with straight lines.</span>
52. Rahzel wants to determine how much gasoline they had every month
=> He used 78 1/3 gallons of gas monthly
=> his wife used 41 3/8 gallons of gas last month
How much is to total of gas that they both used.
First let’s convert this fraction to decimal
=> 78 1/3 = 78.33
=> 41 3/8 = 41.38
Now, let’s start adding.
=> 78.33 + 41.38 = 119.71 this is already rounded to the nearest hundredths.
Answer:
Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.
A. The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.
C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.
D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.
E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake
The answer is E.
1.) y=3
2.) y=13
2y + y = 3y , 22+17= 39
3y = 39 , divide 3 on each side (which cancels out the 3 with the y which leaves you with y)
3.) y=10
3y - y = 2y , 35-15= 20
2y=20 , do the same as # 2.
4.) y=4
9y - 5y = 4y , 48-32= 16
4y=16, do the same as #2 & #3
5.) (4,5)
x + 3y = 19
2x - 3y = -7
solve for x first ; x + 2x= 3x , 19-7 = 12; 3x = 12 ; divide 3 ; x= 4. You know have to make the x's cancel out. so I will multiply -2 on the first equation. -2(x+3y=19)
-2x - 6y = -38
2x - 3y = -7
Now solve for y ; -6y + -3y = -9y , -38 + -7 = -45; -9y=-45 ; divide by -9 (since they are both negative your answer will be positive) ; y= 5
6.) (6,8)
x + 4y = 38
-x - 3y = -30
solve for y first ; 4y - 3y = y , 38-30 = 8; y=8 (that simple!) Now you need to find a common multiple for 4 & 3 which is 12. So you will have to multiply each equation by either 4 or 3.
(x + 4y = 38)*3 = 3x + 12y = 114
(-x - 3y = -30)*4 = -4x - 12y = -120
solve for x ; 3x - 4x = -x , 114-120= -6 ; -x=-6. Since the x has a negative that means there's still a 1 there so -1x=-6 ; you will need to divide this which makes the 6 a positive; x=6
