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2 years ago

I have a question that involves derivatives in the context of a triangle. Can someone confirm that the answer is 39/41?

Mathematics

1 answer:

astra-53 [7]2 years ago

5 0

X^2 + y^2 = z^2
d/dt[ x^2 + y^2 ] = d/dt[ z^2 ]
d/dt[ x^2 ] + d/dt[ y^2 ] = d/dt[ z^2 ]
2x*dx/dt + 2y*dy/dt = 2z*dz/dt
x*dx/dt + y*dy/dt = z*dz/dt
12*(3*dy/dt) + 5*dy/dt = z*1
12*(3*dy/dt) + 5*dy/dt = 13
12*(3*w) + 5*w = 13 ... letting w = dy/dt
36*w + 5*w = 13
41*w = 13
w = 13/41
dy/dt = 13/41
dx/dt = 3*(dy/dt)
dx/dt = 3*(13/41)
dx/dt = 39/41

You have the correct answer. Nice job.

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Nutka1998 [239]

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Thus,

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3 years ago

Mai and Jada are both running on a treadmill. Mai travels 5 miles in 15 minutes. Jada travels 4 miles in 14 minutes. Who ran at

sasho [114]

Answer:

Mai

Step-by-step explanation:

To find the rate, you must determine who is going faster per minute. To do this, you must divide the total number of miles by the total number of minutes.

Mai: 5 miles / 15 minutes = 1/3 miles per minute

Jada: 4 miles / 14 minutes = 2/7 miles per minute

Then, use a common denominator to make calculations easier:

Mai: 1/3 * 7/7

Jada: 2/7 * 3/3

You should get:

Mai: 7/21 miles per minute

Jada: 6/21 miles per minute

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3 years ago

At a university, 60% of the 7,400 students are female. The student newspaper reports the results of a survey of a random sample

omeli [17]

Given Information:

Population mean = p = 60% = 0.60

Population size = N = 7400

Sample size = n = 50

Required Information:

Sample mean = μ = ?

standard deviation = σ = ?

Answer:

Sample mean = μ = 0.60

standard deviation = σ = 0.069

Step-by-step explanation:

We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10

np ≥ 10

50*0.60 ≥ 10

30 ≥ 10 (satisfied)

n(1 - p) ≥ 10

50(1 - 0.60) ≥ 10

50(0.40) ≥ 10

20 ≥ 10 (satisfied)

The mean of the sampling distribution will be same as population mean that is

Sample mean = p = μ = 0.60

The standard deviation for this sampling distribution is given by

$\sigma = \sqrt{\frac{p(1-p)}{n} }$

Where p is the population mean that is proportion of female students and n is the sample size.

$\sigma = \sqrt{\frac{0.60(1-0.60)}{50} }\\\\\sigma = \sqrt{\frac{0.60(0.40)}{50} }\\\\\sigma = \sqrt{\frac{0.24}{50} }\\\\\sigma = \sqrt{0.0048} }\\\\\sigma = 0.069$

Therefore, the standard deviation of the sampling distribution is 0.069.

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