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1 year ago

When expressing sets, you may write "inf" for infinity and "U" for union.

Mathematics

1 answer:

GarryVolchara [31]1 year ago

3 0

Answer:

The complement of the given set in interval notation is $(-\infty,-5]\cup(6,\infty)$ . It can we written as (-inf,5]U(6,inf).

Step-by-step explanation:

The given set in interval notation is

(−5,6]

It means the set is defined as

$A=\{x|x\in R,-5$

If B is a set and U is a universal set, then complement of set B contains the elements of universal set but not the elements of set B.

Here, universal set is R, the set set of all real numbers.

$U=\{x|x\in R\}$

The complement of the given set is

$A^c=U-A$

$A^c=\{x|x\in R,-\infty$

Complement of the given set in interval notation is

$A^c=(-\infty,-5]\cup(6,\infty)$

Therefore the complement of the given set in interval notation is $(-\infty,-5]\cup(6,\infty)$ . It can we written as (-inf,5]U(6,inf).

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A veterinarian does research on the causes of enteroliths, stones that develop in the colon of horses. She suspects that feeding

skad [1K]

Answer: 8

Step-by-step explanation:

The degree of freedom for t-distribution of testing the differnce between the two population mean is given by :-

$df= n_1+n_2-2$

, where $n_1$ = Size of first sample.

$n_1$ = Size of second sample.

As per given ,

Veterinarian selects five horses that are known to have enteroliths and compares the number of flakes of alfalfa they have eaten over a month with the number of flakes eaten by five horses free of enteroliths.

i.e. $n_1=5$ and $n_2=5$

Now , If she calculates a two-sample confidence interval by hand for the difference in the mean number of flakes fed to horses with and without enteroliths the degrees of freedom she should use are

$df= 5+5-2=8$

Hence, the correct answer is 8.

8 0

1 year ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0

1 year ago

What is the slope of the line shown? a -7/9 b 9/7 c 7/9 d -9/7

max2010maxim [7]

Slope = y2 - y1/x2 - x1
6 - -3/1 - -6
=9/7

7 0

2 years ago

3X3X11 Write the number whose prime factorization is given

mixer [17]

the prime factorization is 99

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2 years ago

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Prompt:

Vadim26 [7]

Answer:

Her answer is wrong because she the object cannot hit the ground at negative seconds. She could’ve have used other methods because she used the quadratic formula. The advantages is that it works for every situation. The disadvantages is that it takes longer. She should’ve used a different method.

Step-by-step explanation:

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2 years ago

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