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butalik [34]
2 years ago
11

A cyclist rides her bike at a speed of 30 kilometers per hour. What is this speed in kilometers per minute? How many kilometers

will the cyclist travel in 2
minutes? Do not round your answers,
Mathematics
1 answer:
Nata [24]2 years ago
3 0

Step-by-step explanation:

The answer is mentioned above.

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A) 0.2g

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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
Please help me with this question
san4es73 [151]

Step-by-step explanation:

Given: f'(x) = x^2e^{2x^3} and f(0) = 0

We can solve for f(x) by writing

\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx

Let u = 2x^3

\:\:\:\:du=6x^2dx

Then

\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du

\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k

We know that f(0) = 0 so we can find the value for k:

f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}

Therefore,

\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)

5 0
2 years ago
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