Answer:
The interval will be narrower if the researchers increase the sample size of droplets.
Step-by-step explanation:
The confidence interval can be obtained using the relation :
Xbar ± Margin of error
Margin of Error = Zcrit * s/sqrt(n)
Zcritical = critical vlaue at the specified α - level
n = sample size ; s = standard deviation
Sample size being the denominator, will reduce the overall value of the error margin as we utilize a larger sample.
Hence, the interval becomes narrower as the error margin is reduced which is achieved by employing an increased sample size.
Answer:
3
Step-by-step explanation:
First we need to find 1/4 of a ft which is 0.25
Then we need to convert 0.25 ft to inches which is 3 we get it buy multiplying 12 to the length value
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Answer:
s represents the shorter leg
3s - 9 represents the longer leg
2s + 1 represents the hypotenuse
(2s + 1)2 = s2 + (3s - 9)2 Pythagorean Theorem
4s2 + 4s + 1 = s2 + 9s2 - 54s + 81 square binomial
4s2 + 4s + 1 = 10s2 - 54s + 81 combine like terms
0 = 6s2 - 58s + 80 set to 0
0 =3s2 - 29s + 40 divide both sides by 2
0 = (3s - 5)(s - 8) FOIL
{5/3, 8}
5/3 is not feasible since 3(5/3) - 9 is negative.
-by-step explanation:
Answer:
The volume of the solid is ![1,021m^3](https://tex.z-dn.net/?f=1%2C021m%5E3)
Step-by-step explanation:
1. Cylinder
For find the volume of a cylinder we use the next formula:
![V_{cylinder} = \pi h r^2 = \pi \cdot 9m\cdot (5m)^2 \approx 706.86m^3](https://tex.z-dn.net/?f=V_%7Bcylinder%7D%20%3D%20%5Cpi%20h%20r%5E2%20%3D%20%5Cpi%20%5Ccdot%209m%5Ccdot%20%285m%29%5E2%20%5Capprox%20706.86m%5E3)
Then the volume of the cylinder is equal to ![706.86m^3](https://tex.z-dn.net/?f=706.86m%5E3)
2. Cones
For find the volume of a cone we use the next formula:
![V_{cone} = \pi \frac{h}{3} r^2 = \pi \cdot \frac{6m}{3} \cdot (5m)^2 \approx 157.08m^3](https://tex.z-dn.net/?f=V_%7Bcone%7D%20%3D%20%5Cpi%20%5Cfrac%7Bh%7D%7B3%7D%20r%5E2%20%3D%20%5Cpi%20%5Ccdot%20%5Cfrac%7B6m%7D%7B3%7D%20%5Ccdot%20%285m%29%5E2%20%5Capprox%20157.08m%5E3)
However there are two cones so we have to multiply the volume of one cone for two and we get the total volume for the cones which is ![314.16m^3](https://tex.z-dn.net/?f=314.16m%5E3)
3. Sum of the volumes
Finally we sum the volumes of the cylinder and the cones for get the final result
![V_{cylinder} + V_{cones} = 706.86m^3 + 314.16m^3 \approx 1021m^3](https://tex.z-dn.net/?f=V_%7Bcylinder%7D%20%2B%20V_%7Bcones%7D%20%3D%20706.86m%5E3%20%2B%20314.16m%5E3%20%5Capprox%201021m%5E3)
So approximating the result is ![1021m^3](https://tex.z-dn.net/?f=1021m%5E3)
Problems of this sort are frequently found in physics. If you study calculus or physics you'll learn how to create the equation representing the velocity of an object in flight.
Here, you don't need to calculate velocity, but rather time. Start with this equation:
v = v0 + a t^2, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration due to gravity (denoted by g instead of a), and t is the elapsed time.
You are told that v0 is 15 ft/sec. Set v = to 0, as the ball stops moving for the tiniest instant at the top of its trajectory. Use g = - 32 ft (per second squared).
Then 0 = 15 ft/sec - 32 [ft/(seconds squared)] t.
Solve this for t. This is the time required for the ball to come to a complete stop at the top of its trajectory.
Finally, multiply this time by 2, since the ball begins to fall and returns to its original height.