Let the train's original speed be s. Recall that distance = (speed)(time). Here, 150 miles = s(time), or 150 = s*t. If its original speed is increased by 5 mph, then the time required tomake the trip is 1 hour less than before;
distance = speed times time, so
150 mi = (s+5)(t-1). Let's eliminate t and solve for s:
Since 150 = s*t, t = 150/s. Subbing 150/s for t in the 2nd equation, we get:
150 s
150 mi = (s+5)(150/s - 1), or 150 = (s+5)(------- - ---- )
s s
Mult. both sides by s to elim. the fraction(s):
150s = (s+5)(150 - s)
Then 150s = 150s - s^2 + 750 - 5s, or
0 = -s^2 - 5s + 750
or 0 = s^2 + 5s - 750
Thus, 0 = (s-25)(s+30), and the roots are s=25 and-30. Only a positive original speed makes sense, so the answer is s = 25 mph.
Step-by-step explanation:
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Answer:
skewed to the right
Step-by-step explanation:
The distance from the mean (110) of the remaining four numbers is presented bellow:
It can be observed that the numbers in the first and second quartiles are much closer to the mean than the numbers in the third and fourth quartiles, which means that there is more data on the left side of the distribution. This results in a longer right-tail, or a distribution skewed to the right.
Actual area = (140-yd x 940-yd) = 131,600 square yards.
Estimated area = (100-yd x 850-yd) = 85,000 square yards
Error = (131,600 - 85,000) = 46,600 square yards
Error as a fraction of the actual area = 46,600 / 131,600 = 0.354...
To change a decimal to a percent,
move the decimal point 2 places that way ==> .
0.354... = 35.4.. % (rounded)
Step-by-step explanation:
1 garm = $50.04
7.5 gram = 50.04 × 7.5 = $375.3