Answer:
1. 625,000 J
2. 100 J
4. 5 kg
5. √5 ≈ 2.236 m/s
Step-by-step explanation:
You should be aware that the SI derived units of Joules are equivalent to kg·m²/s².
To reduce confusion between <em>m</em> for mass and m for meters, we'll use an <em>italic m</em> for mass.
In each case, the "find" variable is what's left after we put the numbers into the formula. It is what the question is asking for. The "given" values are the ones in the problem statement and are the values we put into the formula. The formula is the same in every case.
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1. KE = (1/2)<em>m</em>v² = (1/2)(2000 kg)(25 m/s)² = 625,000 kg·m²/s² = 625,000 J
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2. KE = (1/2)<em>m</em>v² = (1/2)(0.5 kg)(20 m/s)² = 100 kg·m²/s² = 100 J
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4. KE = (1/2)<em>m</em>v²
250 J = (1/2)<em>m</em>(10 m/s)² = 50 m²/s²
(250 kg·m²/s²)/(50 m²/s²) = <em>m</em> = 5 kg
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5. KE = (1/2)<em>m</em>v²
2000 kg·m²/s² = (1/2)(800 kg)v²
(2000 kg·m²/s²)/(400 kg) = v² = 5 m²/s²
v = √5 m/s ≈ 2.236 m/s
F(2)=-29-f(1) = -29-(-16)=-13
Y = 2 x + 18 i think thats the answer :D
Answer:
a. 2.28%
b. 30.85%
c. 628.16
d. 474.67
Step-by-step explanation:
For a given value x, the related z-score is computed as z = (x-500)/100.
a. The z-score related to 700 is (700-500)/100 = 2, and P(Z > 2) = 0.0228 (2.28%)
b. The z-score related to 550 is (550-500)/100 = 0.5, and P(Z > 0.5) = 0.3085 (30.85%)
c. We are looking for a value b such that P(Z > b) = 0.1, i.e., b is the 90th quantile of the standard normal distribution, so, b = 1.281552. Therefore, P((X-500)/100 > 1.281552) = 0.1, equivalently P(X > 500 + 100(1.281552)) = 0.1 and the minimun SAT score needed to be in the highest 10% of the population is 628.1552
d. We are looking for a value c such that P(Z > c) = 0.6, i.e., c is the 40th quantile of the standard normal distribution, so, c = -0.2533471. Therefore, P((X-500)/100 > -0.2533471) = 0.6, equivalently P(X > 500 + 100(-0.2533471)), and the minimun SAT score needed to be accepted is 474.6653
Answer:
100+x
Step-by-step explanation:
hope this helps. pls mark brainliest :D
