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Papessa [141]
3 years ago
14

Plot another sin function of 20% higher frequency over the same range.

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
4 0

Step-by-step explanation:

The frequency of sine function is given by the number of periods in a given range. For example:

  • Frequency for sin(x) is 1 in the interval [0,2\pi].

This means that, if we want another sine function with frequency 20% higher, we need that function to have a frequency of 1.2 in the interval [0,2\pi].

To be easier to see we will consider interval [0,10\pi] instead of [0,2\pi]. In this interval sin(x) has 5 periods, therefore our new sine function should have 6 periods.

Finally, as we can see in the graph, the function sin(\frac{6}{5}x ) (in blue) has a frequency 20% higher than sin(x) (in red). This can be easily seen counting the number of periods between 0 and 10\pi for both functions. 5 for sin(x) and 6 for sin(\frac{6}{5} x).

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Ming has 15 quarters, 30 dimes, and 48 nickels. He wants to group his money so that each group had the same number of each coin.
Colt1911 [192]
Find the greatest common factor. In this case it is 3. Divide each group by 3.
15÷3=5 quarters
30÷3=10 dimes
48÷3=16 nickels
Next count them by the coin's value:
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10 × 0.10 = $1
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= $3.05 in each group and there are three groups.

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Pls help asap <br><br> Which quadratic function is represented by the graph?
DiKsa [7]

Answer:

1st option

Step-by-step explanation:

From the graph the zeros are x = - 3 and x = 1 , then the corresponding factors are

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3 years ago
A data set includes 108 body temperatures of healthy adult humans having a mean of 98.3degrees°F and a standard deviation of 0.6
mylen [45]
<h2>Answer with explanation:</h2>

The confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

Given : Sample size : n=106

Sample mean : \overline{x}=98.3^{\circ}\ F

Standard deviation : \sigma= 0.69^{\circ}F

Significance level : \alpha: 1-0.99=0.01

Critical value : z_{\alpha/2}=2.576

Then , 99​% confidence interval estimate of the mean body temperature of all healthy humans will be :-

98.3\pm (2.576)\dfrac{0.69}{\sqrt{106}}\\\\\approx98.3\pm0.173=(98.3-0.173,\ 98.3+0.173)=(98.127,\ 98.473)

Since 98.6 is higher than the upper limit of the confidence interval (98.473), then this suggest that the mean body temperature could be lower then 98.6 degrees.

The best estimate of population mean is always the sample mean. \mu=\overline{x}=98.3^{\circ}\ F

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3 years ago
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