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Mila [183]
2 years ago
12

A particle moves on the hyperbola xy=15 for time t≥0 seconds. At a certain instant, x=3 and dx/dt=6. Which of the following is t

rue about y at this instant?
Mathematics
1 answer:
Aleks [24]2 years ago
4 0

Answer:

y is decreasing by 10units per second

Step-by-step explanation:

Given xy = 15

If x = 3

y = 15/3

y = 5

dx/dt= 6

Differentiating xy = 15 implicitly with respect to t using product rule

x

xdy/dt + ydx/dt = 0

Substitute the given value into the expression and solve for dy/dt as shown:

3dy/dt + 5(6) = 0

3dy/dt + 30 = 0

3dy/dt = -30

Divide both sides by 3

(3dy/dt )/3 = -30/3

dy/dt = -10

This means that y is decreasing by 10units per second (it is decreasing because of the negative sign)

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Brady recorded the number and color of cars in the parking lot. 45% of the cars in the parking lot were white. If Brady counted
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Answer:

180 cars

Step-by-step explanation:

Like you can see in the picture, I like to set up a proportion, so 81 over X and 45 over 100. Then, you can cross multiply, so 81 x 100 = 8100 and then divide, 8100 ÷ 45 = 180, your answer!

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PLEASE ANSWER
mestny [16]
-4 - 1/3=-4.33333333333<span>

-4 - -1=-3

1/3 - 2=</span>-1.66666666667<span>

1/3 - 3 1/2=</span>-3.16666666667<span>

2 - 3 1/2=</span>-1.5<span>

-1 - 3 1/2=</span><span>-4.5</span>
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3 years ago
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Is 0.158 greater than 0.58
Juli2301 [7.4K]
No. 0.158 is less than 0.58
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2 years ago
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1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
Ad libitum [116K]
Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21
5 0
3 years ago
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