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question:Mary is counting the number of minnows in the pond. 40% of the minnows have red bellies, some percentage have green bellies, and the rest have white bellies. If 20 minnows have red bellies and 15 minnows have white bellies, what percentage of the minnows have green bellies?
answer:First, let's find out the total number of minnows in the pond. Since 40% of the minnows have red bellies and there are 20 minnows with red bellies, we can set up the following equation to find the total number of minnows (T): 40% of T = 20 minnows 0.40 * T = 20 T = 20 / 0.40 T = 50 minnows Now we know there are 50 minnows in total. We already know that 20 have red bellies and 15 have white bellies. So, the number of minnows with green bellies would be the total number minus the red-bellied and white-bellied minnows: Number of green-bellied minnows = Total minnows - Red-bellied minnows - White-bellied minnows Number of green-bellied minnows = 50 - 20 - 15 Number of green-bellied minnows = 15 Now, to find the percentage of minnows with green bellies, we divide the number of green-bellied minnows by the total number of minnows and multiply by 100: Percentage of green-bellied minnows = (Number of green-bellied minnows / Total minnows) * 100 Percentage of green-bellied minnows = (15 / 50) * 100 Percentage of green-bellied minnows = 0.30 * 100 Percentage of green-bellied minnows = 30% So, boxed{30%} of the minnows have green bellies.
question:parallelogram EFGH, the measure of angle EFG is twice the measure of angle FGH. Determine the measure of angle EHG.
answer:Given that EFGH is a parallelogram, we know that: 1. angle EFG and angle EHG are opposite angles, hence they are equal. 2. angle EFG + angle FGH = 180^circ because consecutive angles in a parallelogram are supplementary. Since angle EFG = 2angle FGH, let angle FGH = x. Then angle EFG = 2x. Using the supplementary angles property: [ 2x + x = 180^circ ] [ 3x = 180^circ ] [ x = 60^circ ] Therefore, angle FGH = 60^circ and angle EFG = 2 times 60^circ = 120^circ. Since angle EHG is equal to angle EFG (opposite angles in a parallelogram): [ angle EHG = boxed{120^circ} ]
question:Tom's rabbit can run at 25 miles per hour. His cat can run 20 miles per hour. The cat gets a 15minute head start. In hours, how long will it take for the rabbit to catch up?
answer:First, let's determine how far the cat will have run during its 15-minute head start. Since the cat runs at a speed of 20 miles per hour, we need to convert the 15 minutes to hours to calculate the distance. 15 minutes is 15/60 hours, which is 0.25 hours. Now, we can calculate the distance the cat travels in that time: Distance = Speed × Time Distance_cat = 20 miles/hour × 0.25 hours = 5 miles The rabbit needs to cover this 5-mile distance to catch up to the cat. Since the rabbit runs at 25 miles per hour, we can calculate the time it takes for the rabbit to catch up by using the formula: Time = Distance / Speed The rabbit needs to make up the 5-mile difference at a relative speed difference, which is the rabbit's speed minus the cat's speed: Relative speed = Speed_rabbit - Speed_cat Relative speed = 25 miles/hour - 20 miles/hour = 5 miles/hour Now, we can calculate the time it takes for the rabbit to catch up: Time_rabbit = Distance_cat / Relative speed Time_rabbit = 5 miles / 5 miles/hour = 1 hour So, it will take the rabbit boxed{1} hour to catch up to the cat.
question:Let p, p', q, and q' be real numbers with p and p' nonzero. Consider the equations px^2 + q = 0 and p'x^2 + q' = 0. The smallest solution to px^2+q = 0 is less than the smallest solution to p'x^2 + q' = 0 if and only if mathrm{(A) } p'q < pq' qquad mathrm{(B) }pq'<p'q qquad mathrm{(C) } pq<p'q' qquad mathrm{(D) } sqrt{frac{q}{p}}<sqrt{frac{q'}{p'}} qquad mathrm{(E) }sqrt{frac{q'}{p'}}<sqrt{frac{q}{p}} A) p'q < pq' B) pq' < p'q C) pq < p'q' D) sqrt{frac{q}{p}} < sqrt{frac{q'}{p'}} E) sqrt{frac{q'}{p'}} < sqrt{frac{q}{p}}
answer:1. **Identify the solutions to the equations**: Solve the equations: [ px^2 + q = 0 implies x^2 = -frac{q}{p} implies x = pm sqrt{-frac{q}{p}}, ] [ p'x^2 + q' = 0 implies x^2 = -frac{q'}{p'} implies x = pm sqrt{-frac{q'}{p'}}. ] 2. **Set up the inequality condition**: We need the smallest solution (negative square root) of px^2+q=0 to be less than the smallest solution of p'x^2+q'=0: [ -sqrt{-frac{q}{p}} < -sqrt{-frac{q'}{p'}}. ] 3. **Simplify the inequality**: Removing the negative signs and reversing the inequality: [ sqrt{-frac{q}{p}} > sqrt{-frac{q'}{p'}}. ] This simplifies to: [ sqrt{frac{q}{p}} > sqrt{frac{q'}{p'}}, ] or: [ sqrt{frac{q'}{p'}} < sqrt{frac{q}{p}}. ] Conclusion: This corresponds to option text{E}. The final answer is boxed{E}.