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kobusy [5.1K]
4 years ago
6

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

e recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order.
Mathematics
1 answer:
SVEN [57.7K]4 years ago
7 0

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

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Step-by-step explanation:

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3 years ago
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4 years ago
u(x) = x5 – x4 + x2 and v(x) = –x2, which expression is equivalent to (StartFraction u Over v EndFraction) (x)?
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5 0
3 years ago
Read 2 more answers
Will reward brainliest!
Lina20 [59]

Option D: Two irrational solutions

Explanation:

The equation is 17+3 x^{2}=6 x

Subtracting 6x from both sides, we have,

3x^{2} -6x+17=0

Solving the equation using quadratic formula,

x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}

Simplifying the expression, we get,

\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}

Taking out the common terms and simplifying, we have,

\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}

Dividing by 3, we get,

x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}

Hence, the equation has two irrational solutions.

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3 years ago
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Rashid [163]
The correct answer is c have a great day
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3 years ago
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