1/3ton=____lb ..........
2 answers:
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The answer is 
![\sqrt[3]{ \frac{ a^{6} b^{9} }{-64} } = \frac{ \sqrt[3]{a^{6} b^{9}} }{ \sqrt[3]{-64} } = \frac{ \sqrt[3]{a^{6}} * \sqrt[3]{ b^{9}} }{ \sqrt[3]{-64}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B%20a%5E%7B6%7D%20b%5E%7B9%7D%20%20%7D%7B-64%7D%20%7D%20%3D%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7Ba%5E%7B6%7D%20b%5E%7B9%7D%7D%20%7D%7B%20%20%5Csqrt%5B3%5D%7B-64%7D%20%20%7D%20%3D%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7Ba%5E%7B6%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B%20b%5E%7B9%7D%7D%20%7D%7B%20%20%5Csqrt%5B3%5D%7B-64%7D%7D%20)
Since (xᵃ)ᵇ = xᵃ*ᵇ, then:
a⁶ = (a²)³
b⁹ = (b³)³
Also: 64 = 4³
![\frac{ \sqrt[3]{a^{6}} * \sqrt[3]{ b^{9}} }{ \sqrt[3]{-64}}= \frac{ \sqrt[3]{( a^{2})^{3} }*\sqrt[3]{( b^{3})^{3} } }{\sqrt[3]{( -4)^{3} }} = \frac{ a^{2} * b^{3} }{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Csqrt%5B3%5D%7Ba%5E%7B6%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B%20b%5E%7B9%7D%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B-64%7D%7D%3D%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B%28%20a%5E%7B2%7D%29%5E%7B3%7D%20%7D%2A%5Csqrt%5B3%5D%7B%28%20b%5E%7B3%7D%29%5E%7B3%7D%20%7D%20%7D%7B%5Csqrt%5B3%5D%7B%28%20-4%29%5E%7B3%7D%20%7D%7D%20%3D%20%5Cfrac%7B%20a%5E%7B2%7D%20%2A%20b%5E%7B3%7D%20%7D%7B-4%7D%20)
Since (xᵃ)ᵇ = xᵃ*ᵇ, then:
a⁶ = (a²)³
b⁹ = (b³)³
Also: 64 = 4³
Answer:
a) represent the sample mean. And that represent the best estimator for the population mean since
b) The 90% confidence interval is given by (17.3;21.1)
c) No, because 18 is above the lower limit of the confidence interval.
d) Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.
Step-by-step explanation:
1) Notation and definitions
n=17 represent the sample size
Part a
represent the sample mean. And that represent the best estimator for the population mean since
Part b
represent the sample standard deviation
m represent the margin of error
Confidence =90% or 0.90
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 means 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.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
2) Calculate the critical value tc
In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by and
. The degrees of freedom are given by:
We can find the critical values in excel using the following formulas:
"=T.INV(0.05,16)" for
"=T.INV(1-0.05,16)" for
The critical value
3) Calculate the margin of error (m)
The margin of error for the sample mean is given by this formula:
4) Calculate the confidence interval
The interval for the mean is given by this formula:
And calculating the limits we got:
The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)
Part c
No, because 18 is above the lower limit of the confidence interval.
Part d
Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.