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yan [13]
1 year ago
15

Evaluate function at the given value using the remainder theorem.

Mathematics
1 answer:
garik1379 [7]1 year ago
5 0

Answer:

6

Step-by-step explanation:

f(a)=a^4-2a^3 -10a^2 -7a \\\\f(-2) = (-2)^4-2(-2)^3 -10(-2)^2 -7(-2)\\\\~~~~~~~~~=16-2(-8)-10(4)+14\\\\~~~~~~~~~=30+16-40\\\\~~~~~~~~~=46-40\\\\~~~~~~~~~=6

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The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
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Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

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Step-by-step explanation:


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