Construct a table of values for the following functions using the integers from -4 to 4.
a. F(x)=6/x-2
b. r(x)=6x+12/x^-4
Answers
Step-by-step explanation:
Find the table attached
a) Given
F(x) = 6/x-2
When x = -4
F(-4) = 6/-4-2
F(-4) = 6/-6
F(-4) = -1
F(x) = 6/x-2
When x = -3
F(-3) = 6/-3-2
F(-3) = 6/-5
F(-3) = -1.2
F(x) = 6/x-2
When x = -2
F(-2) = 6/-2-2
F(-2) = 6/-4
F(-2) = -1.5
F(x) = 6/x-2
When x = -1
F(-1) = 6/-1-2
F(-1) = 6/-3
F(-1) = -2.0
F(x) = 6/x-2
When x = 0
F(0) = 6/0-2
F(0) = 6/-2
F(0) = -3
F(x) = 6/x-2
When x = 1
F(1) = 6/1-2
F(1) = 6/-1
F(1) = -6
F(x) = 6/x-2
When x = 2
F(2) = 6/2-2
F(2) = 6/0
F(2) = infty
F(x) = 6/x-2
When x = 3
F(3) = 6/3-2
F(3) = 6/1
F(3) = 6
F(x) = 6/x-2
When x = 4
F(4) = 6/4-2
F(4) = 6/2
F(4) = 3
b) Given
r(x)=6x+12/x^-4
When x = -4
r(-4) = 6(-4)+12/(-4)^-4
r(-4) = -24+12/(1/256)
r(-4) = -12(256)
r(-4) = -3072
When x = -3
r(-3) = 6(-3)+12/(-3)^-4
r(-3) = -18+12/(1/81)
r(-3) = -6(81)
r(-3) = -486
When x = -2
r(-2) = 6(-2)+12/(-2)^-4
r(-2) = -12+12/(1/16)
r(-2) = -0(16)
r(-2) = 0
When x = -1
r(-1) = 6(-1)+12/(-1)^-4
r(-1) = -6+12/(1)
r(-1) = -6+12
r(-1) = 6
When x = 0
r(0) = 6(0)+12/(0)^-4
r(0) = 0+12/0
r(0) = 12/0
r(0) = infty
When x = 1
r(1) = 6(1)+12/(1)^-4
r(1) = 6+12/1
r(1) = 18(1)
r(1) = 18
When x = 2
r(2) = 6(2)+12/(2)^-4
r(2) = 12+12/1/16
r(2) = 24(16)
r(2) = 384
When x = 3
r(3) = 6(3)+12/(3)^-4
r(3) = 18+12/1/81
r(3) = 30(81)
r(3) = 2430
When x = 4
r(4) = 6(4)+12/(4)^-4
r(4) = 24+12/1/256
r(4) = 36(256)
r(4) = 9216
We want to construct tables of values for the two given functions.
The tables are:
a)
[tex]\left[\begin{array}{ccc}x&y\\-4&-7/2\\-3&-4\\-2&-5\\-1&-8\\0&NaN\\1&4\\2&1\\3&0\\4&-1/2\end{array}\right][/tex]
b)
[tex]\left[\begin{array}{ccc}x&y\\-4&3,048\\-3&954\\-2&180\\-1&6\\0&0\\1&18\\2&204\\3&990\\4&3,096\end{array}\right][/tex]
A table will be something like:
[tex]\left[\begin{array}{ccc}x&y\\-4&\\-3&\\-2&\\-1&\\0&\\1&\\2&\\3&\\4&\end{array}\right][/tex]
Where the values of x go from -4 to 4.
To complete the tables, we just need to evaluate the functions in each one of the x-values at the left, and the outcome will be placed at the right.
a) f(x) = 6/x - 2
Now we just need to evaluate the function in all the given points:
f(-4) = 6/(-4) - 2 = -3/2 - 4/2 = -7/2
f(-3) = 6/-3 - 2 = -4
f(-2) = 6/-2 - 2 = -5
f(-1) = 6/-1 - 2 = -8
f(0) is undefined, as we can't divide by zero, here we can write NaN (Not a number).
f(1) = 6/1 - 2 = 4
f(2) = 6/2 - 2 = 1
f(3) = 6/3 - 2 = 0
f(4) = 6/4 - 2 = -1/2
Now we put all of these in the correspondent place of the table:
[tex]\left[\begin{array}{ccc}x&y\\-4&-7/2\\-3&-4\\-2&-5\\-1&-8\\0&NaN\\1&4\\2&1\\3&0\\4&-1/2\end{array}\right][/tex]
b) We do the same thing, here we have:
r(x) = 6*x + 12/x^-4 = 6*x + 12*x^4
Now we evaluate this in the given values:
r(-4) = 6*(-4) + 12*(-4)^4 = 3,048
r(3) = 6*(-3) + 12*(-3)^4 = 954
r(-2) = 6*(-2) + 12*(-2)^4 = 180
r(-1) = 6*(-1) + 12*(-1)^4 = 6
r(0) = 6*0 + 120^4 = 0
r(1) = 6*1 + 12*1^4 = 18
r(2) = 6*2 + 12*2^4 = 204
r(3) = 6*3 + 12*3^4 = 990
r(4) = 6*4 + 12*4^4 = 3,096
Now we place these values in the correspondent place on the table:
[tex]\left[\begin{array}{ccc}x&y\\-4&3,048\\-3&954\\-2&180\\-1&6\\0&0\\1&18\\2&204\\3&990\\4&3,096\end{array}\right][/tex]
These are our two tables.
If you want to learn more, you can read.
https://brainly.com/question/8629807
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Answers
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Step-by-step explanation:
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1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following situations:
a) H1:μ>μ0,z=1.88
b) H1:μ<μ0,z=−2.75
c) H1:μ≠μ0,z=2.88
2. Let the test statistics T have t distribution when H0 is true. Find the p-value for each of the following situations (provide an interval if the exact one cannot be found using a table):
a) H1:μ>μ0,n=16,t=3.733
b) H1:μ<μ0,df=23,t=−2.500
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Answers
Answer:
1 a [tex]p -value = 0.030054[/tex]
1b [tex]p -value = 0.0029798[/tex]
1c [tex]p -value = 0.0039768[/tex]
2a [tex]p-value = 0.00099966[/tex]
2b [tex]p-value = 0.00999706[/tex]
2c [tex]p-value = 0.0654412[/tex]
Step-by-step explanation:
Considering question a
The alternative hypothesis is H1:μ>μ0
The test statistics is z =1.88
Generally from the z-table the probability of z =1.88 for a right tailed test is
[tex]p -value = P(Z > 1.88) = 0.030054[/tex]
Considering question b
The alternative hypothesis is H1:μ<μ0
The test statistics is z=−2.75
Generally from the z-table the probability of z=−2.75 for a left tailed test is
[tex]p -value = P(Z < -2.75) = 0.0029798[/tex]
Considering question c
The alternative hypothesis is H1:μ≠μ0
The test statistics is z=2.88
Generally from the z-table the probability of z=2.88 for a right tailed test is
[tex]p -value = P(Z >2.88) = 0.0019884[/tex]
Generally the p-value for the two-tailed test is
[tex]p -value = 2 * P(Z >2.88) = 2 * 0.0019884[/tex]
=> [tex]p -value = 0.0039768[/tex]
Considering question 2a
The alternative hypothesis is H1:μ>μ0
The sample size is n=16
The test statistic is t = 3.733
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
=> [tex]df = 16 - 1[/tex]
=> [tex]df = 15[/tex]
Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of [tex]df = 15[/tex] for a right tailed test is
[tex]p-value = t_{3.733 , 15} = 0.00099966[/tex]
Considering question 2b
The alternative hypothesis is H1:μ<μ0
The degree of freedom is df=23
The test statistic is ,t= −2.500
Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is
[tex]p-value = t_{-2.500 , 23} = 0.00999706[/tex]
Considering question 2c
The alternative hypothesis is H1:μ≠μ0
The sample size is n= 7
The test statistic is ,t= −2.2500
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
=> [tex]df = 7 - 1[/tex]
=> [tex]df = 6[/tex]
Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of [tex]df = 6[/tex] for a left tailed test is
[tex]t_{-2.2500 , 6} = 0.03272060[/tex]
Generally the p-value for t= −2.2500 for a two tailed test is
[tex]p-value = 2 * 0.03272060 = 0.0654412[/tex]
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Answers
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Answers
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Answers
Answer:
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Step-by-step explanation:
Answer:
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Step-by-step explanation:
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Answers
9514 1404 393
Answer:
96 cubic feet
Step-by-step explanation:
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V = LWH
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Answers
Answer:
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Step-by-step explanation:
You never showed a picture, but im pretty sure this is correct
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Answers
Answer:
Step-by-step explanation:
A larger set that a particular set can be a subset of is a Universal set. A universal set is parent where all other sets are derived from.
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Example: Find all the subsets of set A = {1,2,3,4}
Solution: Given, A = {1,2,3,4}
Subsets =
The subsets of any set consisting of all possible sets including its elements and the null set. Let us understand with the help of an example.
Example: Find all the subsets of set A = {1,2,3,4}
{}
{1}, {2}, {3}, {4},
{1,2}, {1,3}, {1,4}, {2,3},{2,4}, {3,4},
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[tex]A = \{1, 3, 5\}[/tex]
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Answers
Answer:
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Explanation:
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The mean and standard deviation of a random sample of n measurements are equal to and , respectively. a. Find a % confidence interval for if n. b. Find a % confidence interval for if n. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]33.55 < \mu < 35.5[/tex]
b
[tex]34.03 < \mu < 34.969 [/tex]
c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
d
The correct option is E
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 34.5[/tex]
The standard deviation is [tex]s = 3.4[/tex]
Generally given that the confidence level is 95% then the level of significance is
[tex]\alpha = (100 - 95)\%[/tex]
=> [tex]\alpha = 0.05 [/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Considering question a
From the question n = 49
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{49} }[/tex]
=> [tex]E = 0.952 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.952 < p < 34.5 + 0.952[/tex]
=> [tex]33.55 < \mu < 35.5[/tex]
Considering question b
From the question n = 196
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{196} }[/tex]
=> [tex]E = 0.4687 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.4687 < p < 34.5 +0.4687[/tex]
=> [tex]34.03 < \mu < 34.969 [/tex]
Considering question c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
Now when the sample size is quadrupled i.e from n = 49 to n = 196
The width of the confidence interval decrease by 2 from 1.904 to 0.9374
At a school, there are 120 athletes. The ratio of boy athletes to giri athletes is 3:5. How many of the athletes are girls?
А
75
B
45
с
24
D 5
Answers
Answer:
a
Step-by-step explanation:
because there are over half girls and half of 120 is 60 and 75 is over half
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Answer:
157
Step-by-step explanation:
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Step-by-step explanation:
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Answer:
x=3
Step-by-step explanation:
-5 +10
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Solve: x^2=10
A) x= +5
B) x= +100
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Answers
Answer:
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Step-by-step explanation:
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Question Help When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 36 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 3000 batteries, and 3% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?
Answers
Answer:
The probability is [tex]P(X \le 2 ) = 0.9072[/tex]
The company will accept 90.72% of the shipment and will reject [tex](100 -90.72) = 9.2\%[/tex] of the shipment , so many of the shipment are rejected
Step-by-step explanation:
From the question we are told that
The sample size is n = 36
The proportion that did not meet the requirement is [tex]p = 0.03[/tex]
Generally the probability that the whole shipment is accepted is equivalent to the probability that there is at most 2 batteries that do not meet the requirement , this is mathematically represented as
[tex]P(X \le 2 ) = [ P(X = 0 ) + P(X = 1 ) + P(X = 0)][/tex]
=> [tex]P(X \le 2 ) = [ [^{n}C_0 * (p)^{0} *(1-p)^{n-0} ] + [^{n}C_1 * (p)^{1} *(1-p)^{n-1} ] + [^{n}C_2 * (p)^{2} *(1-p)^{n-2} ]][/tex]
Here C stands for Combination (so we will be making the combination function in our calculators )
So
=> [tex]P(X \le 2 ) = [ [^{36}C_0 * (0.03)^{0} *(1-0.03)^{36-0} ] + [^{36}C_1 * (0.03)^{1} *(1-0.03)^{36-1} ] + [^{36}C_2 * (0.03)^{2} *(1-0.03)^{36-2} ]][/tex]
=> [tex]P(X \le 2 ) = [ [1 * 1 * 0.3340 ] + [36* 0.03 *0.3444 ] + [630 * 0.0009 *(0.355 ]][/tex]
=>[tex]P(X \le 2 ) = 0.9072[/tex]
The company will accept 90.72% of the shipment and will reject [tex](100 -90.72) = 9.2\%[/tex] of the shipment , so many of the shipment are rejected