You have fit a regression model with two regressors to a data set that has 20 observations. The total sum of squares is 1000 and the model (regression) sum of squares is 750. What is the adjusted R-squared value for this model
Answer:
Hence the adjusted R-squared value for this model is 0.7205.
Step-by-step explanation:
Given n= sample size=20
Total Sum of square (SST) =1000
Model sum of square(SSR) =750
Residual Sum of Square (SSE)=250
The value of R ^2 for this model is,
R^2 = \frac{SSR}{SST}
R^2 = 750/1000 =0.75
Adjusted [tex]R^2[/tex] :
[tex]Adjusted R^2 =1- \frac{(1-R^2)\times(n-1)}{(n-k-1)}[/tex]
Where k= number of regressors in the model.
[tex]Adjusted R^2 =1-(19\times 0.25/((20-2-1)) = 0.7205[/tex]
Assume a random variable representing the amount of time it takes for a customer service representative to pick up has a uniform distribution between 15 and 20 minutes. What is the probability that a randomly selected application from this distribution took less than 18 minutes
Answer:
0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
Uniform distribution between 15 and 20 minutes.
This means that [tex]a = 15, b = 20[/tex]
What is the probability that a randomly selected application from this distribution took less than 18 minutes?
[tex]P(X < 18) = \frac{18 - 15}{20 - 15} = 0.6[/tex]
0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.
Can someone help me? I don’t know how to solve the rest. I am struggling and I would be so happy if any of you helped me. Thank you for your help!
solve x^3-7x^2+7x+15
Step-by-step explanation:
\underline{\textsf{Given:}}
Given:
\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx
3
+7x
2
+7x−15
\underline{\textsf{To find:}}
To find:
\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx
3
+7x
2
+7x−15
\underline{\textsf{Solution:}}
Solution:
\textsf{Factor theorem:}Factor theorem:
\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}
(x−a)isafactorP(x)⟺P(a)=0
\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x
3
+7x
2
+7x−15
\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0
\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)
\mathsf{When\;x=-3}Whenx=−3
\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)
3
+7(−3)
2
+7(−3)−15
\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15
\mathsf{P(-3)=63-63}P(−3)=63−63
\mathsf{P(-3)=0}P(−3)=0
\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor
\mathsf{When\;x=-5}Whenx=−5
\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)
3
+7(−5)
2
+7(−5)−15
\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15
\mathsf{P(-5)=175-175}P(−5)=175−175
\mathsf{P(-5)=0}P(−5)=0
\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor
\underline{\textsf{Answer:}}
Answer:
\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x
3
+7x
2
+7x−15=(x−1)(x+3)(x+5)
\underline{\textsf{Find more:}}
Find more:
Using the following equation, find the center and radius: x2 −2x + y2 − 6y = 26 (5 points)
Answer:
Center: (1,3)
Radius: 6
Step-by-step explanation:
Hi there!
[tex]x^2-2x + y^2 - 6y = 26[/tex]
Typically, the equation of a circle would be in the form [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.
To get the given equation [tex]x^2-2x + y^2 - 6y = 26[/tex] into this form, we must complete the square for both x and y.
1) Complete the square for x
Let's take a look at this part of the equation:
[tex]x^2-2x[/tex]
To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:
[tex]x^2-2x+1[/tex]
Great! Now, let's add this to our original equation:
[tex]x^2-2x+1+y^2-6y = 26[/tex]
We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:
[tex]x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27[/tex]
Complete the square:
[tex](x-1)^2+y^2-6y = 27[/tex]
2) Complete the square for y
Let's take a look at this part of the equation [tex](x-1)^2+y^2-6y = 27[/tex]:
[tex]y^2-6y[/tex]
To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:
[tex]y^2-6y+9[/tex]
Great! Now, back to our original equation:
[tex](x-1)^2+y^2-6y+9= 27[/tex]
Remember to add 9 on the other side as well:
[tex](x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36[/tex]
Complete the square:
[tex](x-1)^2+(y-3)^2= 36[/tex]
3) Determine the center and the radius
[tex](x-1)^2+(y-3)^2= 36[/tex]
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.
I hope this helps!
Answer:
[tex]\sqrt{g^2+f^2-c}[/tex]
[tex]g=-1,f=-3,c=-26[/tex]
so, the Center of the equation is [tex](1,3)[/tex]
Center → (1 , 3)[tex]\sqrt{(-1)^2+(-3)^2-(-26})[/tex]
[tex]=\sqrt{1+9+26}[/tex]
[tex]=\sqrt{36}[/tex]
[tex]=6[/tex]
Radius → 6OAmalOHopeO
o the area of a rhombus is 24m²
and one of its diagonals 18cm find
the side of the rhombus
Area of rhombus = 1/2 × d1 × d2
Let the other diagonal be x
ATQ
1/2 × 18 × x = 24
9 × x = 24
x = 24/9
x = 8/3
Now half each diagonal = 9 and 4/3
Now side = b² + p² = h²
9²+(4/3)² = h²
81 + 16/9 = h²
729/9 + 16/9 = h²
745/9 = h²
√(745/9) = h
Therefore the side of the rhombus is √(745/9)cm
Answered by Gauthmath must click thanks and mark brainliest
Please help i need the answer asap!!!
if you know the answer please give it to me as soon as you can!!
Answer:
Choice b.Step-by-step explanation:
Replace (x, y) with (1, 1) and verify if the equations are correct.
You would ignore the x and y in the equations. Testing one of each pair.
a. 3 + 2 = 3, incorrectb. 7 + 2 = 9, correctc. 8 + 1 = 7, incorrectd. 8 - 2 = 4, incorrectIt is obvious that only b. is correct.
Form a polynomial whose zeros and degree are given,
Zeros: - 2, 2,7; degree: 3
Type a polynomial with integer coefficients and a leaning coefficient of 1 in the box below.
F(x)=
Answer:
if the zeros are x = -2, x = 0, and x = 1
then (x + 2), x and (x - 1) are factors of the polynomial
multiply these factors together
p(x) = x(x + 2)(x - 1)
p(x) = x(x2 + x - 2)
p(x) = x3 + x2 - 2x
this is a polynomial of degree 3 with the given roots
Smart phone: Among 239 cell phone owners aged 18-24 surveyed, 103 said their phone was an Android phone. Part: 0 / 30 of 3 Parts Complete Part 1 of 3 (a) Find a point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places. The point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone is .
Answer:
The point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone is 0.4137.
Step-by-step explanation:
The point estimate is the sample proportion.
Sample proportion:
103 out of 249, so:
[tex]p = \frac{103}{249} = 0.4137[/tex]
The point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone is 0.4137.
Why does cube root 7 equal 7 to the 1/3 power
Answer:
Step-by-step explanation:
Here's how you convert:
[tex]\sqrt[n]{x^m}=x^{\frac{m}{n}[/tex] The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.
A couple of examples:
[tex]\sqrt[3]{x^4}=x^{\frac{4}{3}[/tex]
[tex]\sqrt[5]{x^7}=x^{\frac{7}{5}[/tex]
It's that simple. For your problem in particular:
[tex]\sqrt[3]{7}[/tex] is the exact same thing as [tex]\sqrt[3]{7^1}=7^{\frac{1}{3}[/tex]
What is the measure of the angle in red?
70°
Answer:
See pic below.
Answer: C. 380°
Step-by-step explanation:
A building 51 feet tall casts a shadow 48 feet long. Simultaneously, a nearby statue casts a shadow of 16 feet. How tall is the statue? Choose an answer
Answer: 17 feet
Step-by-step explanation:
51/48 = x/16
(51)(16)/48
The statute is 17 feet tall.
What are the similar triangles?Similar triangles are the triangles that have the same shape, but different sizes. The corresponding angles are congruent and the sides are in proportion.
What is the ratio of any two corresponding sides of similar triangles?The ratio of any corresponding sides in two equiangular triangles is always the same.
Let's visualize the situation according to the given question.
AB is the building ,whose height is 51f
BC is the shadow of the building AB, whose length is 48ft.
QR is the shadow of the tower statue, whose length is 16feet.
Let the height of the statue PR be h feet.
In triangle ACB and triangle PRQ
∠ACB = ∠PRQ = 90 degrees
( the objects and shadows are perpendicular to each other)
∠BAC = ∠QPR
( sunray falls on the pole and tower at the same angle, at the same time )
⇒ΔACB similar to ΔPRQ ( AA criterion)
Therefore, the ratio of any two corresponding sides in equiangular triangles is always same.
⇒ AC/CB = PR/RQ
⇒[tex]\frac{51}{48} =\frac{h}{16}[/tex]
⇒ h = [tex]\frac{(51)(16)}{48}[/tex]
⇒ h = 17 feet.
Hence, the statute is 17 feet tall.
Learn more about the similar triangle here:
brainly.com/question/25882965
#SPJ2
Which of the following conversions is possible?
Meters to Liters
Feet to Miles
Pounds to Inches
Grams to Centimeters
Answer:
Feet to Miles
Step-by-step explanation:
Answer:
Feet to Miles
Step-by-step explanation:
Both units must measure the same quantity.
The posted weight limit for a wooden
bridge is 6,500 pounds. A delivery truck
is loaded with identical boxes of canned
goods that weigh 16 pounds each. If the
combined weight of the empty delivery
truck and the driver is 3,512 pounds,
what is the maximum number of boxes
that would keep the combined weight of
the truck, driver, and boxes below the
posted weight limit?
9514 1404 393
Answer:
186
Step-by-step explanation:
Let b represent the number of boxes in the truck. Then for the weight limit to be met, we require ...
3512 +16b < 6500
16b < 2988
b < 186.75
The maximum number of boxes is 186.
The following are on a parabola defining the edge of a ski
(-4, 1), (-2, 0.94), (0.1)
The general form for the equation of a parabola is:
Ax^2+ Bx +C= y
Required:
a. Use the x- and y-values of 1 of the points to build a linear equation with 3 variables: A, B, and C.
b. Repeat this process with 1 of the other to build a 2nd linear equation.
c. Record your equation here. Repeat this process with the other point to build a 3rd equation.
9514 1404 393
Answer:
a) 16A -4B +C = 1
b) 4A -2B +C = 0.94
c) C = 1
Step-by-step explanation:
Substitute the x- and y-values into the general form equation.
a. A(-4)² +B(-4) +C = 1
16A -4B +C = 1
__
b. A(-2)² +B(-2) +C = 0.94
4A -2B +C = 0.94
__
c. A(0)² +B(0) +C = 1
C = 1
_____
Additional comment
Solving these equations gives A=0.015, B=0.06, C=1. The quadratic is ...
0.015x² +0.06x +1 = y
I need help answering this ASAP
Answer:
"D"
if you multiply by Conjugate
the denominator would end up A^2 - b^2
the answer has 25 - 10x
that is D
Step-by-step explanation:
log_c(A)=2
log_c(B)=5,
solve log_c(A^5B^3)
We know that [tex]\log_a(bc)=\log_a(b)+\log_a(c)[/tex].
Using this rule,
[tex]\log_c(A^5B^3)=\log_c(A^5)+\log_c(B^3)[/tex].
We also know that [tex]\log_c(a^b)=b\log_c(a)[/tex].
Using this rule,
[tex]\log_c(A^5)+\log_c(B^3)=5\log_c(A)+3\log_c(B)[/tex]
Now we know that [tex]\log_c(A)=2,\log_c(B)=5[/tex] so,
[tex]5\cdot2+3\cdot5=10+15=\boxed{25}[/tex].
Hope this helps :)
Help. The graph shows the system of equations below.
2x -3y = -6
y = - 1/3x -4
9514 1404 393
Answer:
(a) The blue line ... solution ... (-6, -2)..
Step-by-step explanation:
The second equation describes a line with negative slope and a y-intercept of -4. This is clearly the red line on the graph.
The blue line represents the equation 2x -3y = -6.
The point of intersection of the two lines is (-6, -2), so that is the solution to the system of equations. This, by itself, is sufficient for you to choose the correct answer.
Which is the
Simplified form
r-7+s-12
Answer:
r + s - 19
General Formulas and Concepts:
Algebra I
Terms/CoefficientsStep-by-step explanation:
Step 1: Define
Identify
r - 7 + s - 12
Step 2: Simplify
Combine like terms [constants]: r + s - 19Oh Brian~
I need help again
Answer:
18c^3d^9
Step-by-step explanation:
2c^3 d^2 * 9d^7
We know that we add the exponents when the base is the same
2*9 c^3 d^(2+7)
18c^3d^9
What is (x+13)^2? pls help!!!
!!!!PLEASE HELP!!!!
What is the following quotient?
2
√13+ /11
O 13-2./11
13+ 11
6
13+ V11
12
√13 - 1
(D)
Step-by-step explanation:
Multiply and divide the fraction by the conjugate:
[tex]\dfrac{2}{\sqrt{13}+\sqrt{11}}×\dfrac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}[/tex]
[tex]= \dfrac{2(\sqrt{13} - \sqrt{11})}{(\sqrt{13})^2 - (\sqrt{11})^2}[/tex]
[tex]=\dfrac{2(\sqrt{13} - \sqrt{11})}{2}[/tex]
[tex]=\sqrt{13} - \sqrt{11}[/tex]
Use formula autocomplete to enter a sum function in cell B7 to calculate the total of cells in B2:B6
Excel enables the users to perform mathematics basic and advanced function with just one formula.
The formula for sum of entire row or column can be done with just entering a single formula and results are shown in seconds.
The formula for sum of few column cells is,
=SUM(B2:B6)
The spreadsheet allows the user to enter various formula and results are displayed withing seconds.
There are formulas for basic math functions and there are also formulas for advance mathematics calculations. For addition of values of many cells sum formula is used and range is assigned for reference.
The formula adds all the values of selected cells and displays the results in different cell.
Learn more at https://brainly.com/question/24365931
Which expression is equivalent to 3/2
Answer:
C
Step-by-step explanation:
Suppose that a random sample of size 64 is to be selected from a population with mean 50 and standard deviation 5. (a) What are the mean and standard deviation of the sampling distribution of x
Answer:
Mean of sampling distribution = 50
Standard deviation of sampling distribution, = 0.625
Step-by-step explanation:
Given :
Mean, μ = 50
Standard deviation, σ = 5
Sample size, n = 64
The mean of sampling distribution, μxbar = population mean, μ
μxbar = μ
According to the central limit theorem, the sampling distribution converges to the population mean as the sample size increases, hence, , μxbar = μ = 50
Standard deviation of sampling distribution, σxbar = σ/√n
σxbar = 5/√64 = 5 / 8 = 0.625
I need to know the answer ASAP
Answer:
Step-by-step explanation:
The graph shows the distribution of the number of text messages young adults send per day. The distribution is approximately Normal, with a mean of 128 messages and a standard deviation of 30 messages.
What percentage of young adults send more than 158 text messages per day?
16%
34%
68%
84%
Answer:
You'd think its 34% but apparently it's 16%.
I hope this is right. If its not then it must be 34%.
(A) -- or maybe (B). 80% confident it is A.
ED2021
x^{2} +y^{2} =?
cho mình hỏi với
Answer:
[tex]{ \sf{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy }}[/tex]
Step-by-step explanation:
[tex]{ \tt{ {(x + y)}^{2} = (x + y)(x + y) }} \\ { \tt{ {(x + y)}^{2} = ( {x}^{2} + 2xy + {y}^{2}) }} \\ { \tt{( {x}^{2} + {y}^{2} ) = {(x + y)}^{2} - 2xy}}[/tex]
An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from soil. Out of 155 seeds planted in soil containing 3% mushroom compost by weight, 74 germinated. Out of 155 seeds planted in soil containing 5% mushroom compost by weight, 86 germinated. Can you conclude that the proportion of seeds that germinate differs with the percent of mushroom compost in the soil
Solution :
Let [tex]p_1[/tex] and [tex]p_2[/tex] represents the proportions of the seeds which germinate among the seeds planted in the soil containing [tex]3\%[/tex] and [tex]5\%[/tex] mushroom compost by weight respectively.
To test the null hypothesis [tex]H_0: p_1=p_2[/tex] against the alternate hypothesis [tex]H_1:p_1 \neq p_2[/tex] .
Let [tex]\hat p_1, \hat p_2[/tex] denotes the respective sample proportions and the [tex]n_1, n_2[/tex] represents the sample size respectively.
[tex]$\hat p_1 = \frac{74}{155} = 0.477419[/tex]
[tex]n_1=155[/tex]
[tex]$p_2=\frac{86}{155}=0.554839[/tex]
[tex]n_2=155[/tex]
The test statistic can be written as :
[tex]$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}[/tex]
which under [tex]H_0[/tex] follows the standard normal distribution.
We reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance, if the P-value [tex]<0.05[/tex] or if [tex]|z_{obs}|>Z_{0.025}[/tex]
Now, the value of the test statistics = -1.368928
The critical value = [tex]\pm 1.959964[/tex]
P-value = [tex]$P(|z|> z_{obs})= 2 \times P(z< -1.367928)$[/tex]
[tex]$=2 \times 0.085667$[/tex]
= 0.171335
Since the p-value > 0.05 and [tex]$|z_{obs}| \ngtr z_{critical} = 1.959964$[/tex], so we fail to reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the [tex]\text{proportion}[/tex] of the seeds that [tex]\text{germinate differs}[/tex] with the percent of the [tex]\text{mushroom compost}[/tex] in the soil.
Please help me with this question
