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Rudiy27
3 years ago
13

A chord consists of notes that sound good together. The C major chord, starting at middle C, has the following

Mathematics
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

D. 3

Step-by-step explanation:

I mathed it

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Right circular cone

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A≈703.72cm²

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Answer:

Step-by-step explanation: their go your answer plz let me know got right then so plz rate me the  most brainlest preson

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Write an algebrale expression for the phrase "elghe more than three times a<br> number
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x 3

Step-by-step explanation:

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Simplify 3/5(10g-5k)-(-3g+2k)
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3 years ago
The sugar content of the syrup in canned peaches is normally distributed. A random sample of n = 10 cans yields a sample standar
Serggg [28]

Answer:

The 95% confidence interval is given by:

3.30<σ<8.76

Step-by-step explanation:

1) Data given and notation

s=4.798 represent the sample standard deviation

\bar x represent the sample mean

n=10 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,9)" "=CHISQ.INV(0.975,9)". so for this case the critical values are:

\chi^2_{\alpha/2}=19.022

\chi^2_{1- \alpha/2}=2.700

And replacing into the formula for the interval we got:

\frac{(9)(4.798)^2}{19.022} \leq \sigma \frac{(9)(4.798)^2}{2.700}

10.892 \leq \sigma^2 \leq 76.736

Now we just take square root on both sides of the interval and we got:

3.30 \leq \sigma \leq 8.76

So the best option would be:

3.30<σ<8.76

7 0
2 years ago
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