Answer:
They are 260 miles apart. No it would not be possible.
Step-by-step explanation:
If they are 260 miles apart and you drive at a constant speed on 70mph. You will have to 260/70 which = 3.7 or 4 so it would not be possible.
9514 1404 393
Answer:
a, c, b
Step-by-step explanation:
Collect terms, subtract the constant, divide by the y-coefficient.
(a) 4y + (y - 1) = 29 ⇒ y = 6
(c) (2y + 3) - 4 = 9 ⇒ y = 5
(b) 4y - y + 1 = 13 ⇒ y = 4 . . . . . this equation is not what you have written
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<em>Additional comment</em>
The second equation suffers from a typo, so your answer may vary.
Answer: 125/242
Step-by-step explanation: Probability is favorable outcomes over all outcome, since there were 250+234 total outcome, and 250 were favorable (girls), then the probability is 250/250+234=250/484=125/242
0.25 is going to be the unit rate when -x in the equation is considered to be 1
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
