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madreJ [45]
2 years ago
9

. Waipio Rd in Honokaa, Hawaii, is one of the steepest roads in the world. It rises 450 ft over a horizontal distance of 1,000 f

t. a. What is the percentage grade of this hill? Round to the nearest tenth of a percent.
Mathematics
1 answer:
Deffense [45]2 years ago
8 0

Answer: 50%

Step-by-step explanation:

From the question, we're given the information that Waipio Rd in Honokaa, Hawaii, is one of the steepest roads in the world and that it rises 450 ft over a horizontal distance of 1,000 ft.

Therefore, the percentage grade of this hill will be calculated as the height over the length and this will be:

= 450/1000 × 100

= 45%

= 0.45

= 0.5 to nearest tenth

= 50% to nearest tenth.

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Frank had hip replacement surgery and was given a prescription with instructions to take a 200 milligram (mg) tablet three times
IgorC [24]

Answer:

12,600mg per 21 days

Step-by-step explanation:

200mg three times a day → (200mg)(3) = 600mg per a day

21 days → (21)(600mg) = 12,600mg per 21 days

5 0
2 years ago
The ratio of the radii of two similar cylinders is 3:5. The volume of the smaller cylinder is 54pi cubic centimeters. Find the v
8090 [49]
5/3 = 1.66667

1.666666^3=4.6296 (Volume is measured in cubic units)
54x4.6396=250
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8 0
2 years ago
Find square root of 0.0324
irina [24]

Answer:

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Step-by-step explanation:

6 0
1 year ago
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
Daniel’s tennis team made $582.85 from running the concession stand at a basketball game. The food and drinks cost the team $64.
Kobotan [32]
Let x be the number of members in Daniel's tennis team.

The amount made = $582.85
Expenses (food and drinks) = $64
Balance remaining for sharing = 582.85 - 64 = $518.85

Amount received by each player = (amount made - Expenses)/Number of members = 518.85/x
7 0
2 years ago
Read 2 more answers
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