PLEASE HELP!! emir learns to perform 2 vocal pieces during each week of lessons. how many weeks of lessons will emir need before
1 answer:
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Answer:
A. 4".
B. Approximately 9.54".
C. Approximately 13.54".
Step-by-step explanation:
Please find the attachment.
Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.
We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.
A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.
We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.
We can find the value of x using Pythagorean theorem as:
Upon taking square root of both sides of our equation we will get,
Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.
B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.
We can find the value of y using Pythagorean theorem as:
Upon substituting our given values we will get,
Upon taking square root of both sides of our equation we will get,
Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.
C. We can see from our diagram that the length of longer diagram will be the sum of x and y.
Therefore, the length of longer diagonal is approximately 13.54 inches.
Answer:
Step-by-step explanation:
Hello!
Given the data for the variables:
Y: Selling price of a house on the shore of Tawas Bay
X₁: Number of bathrooms of a house on the shore of Tawas Bay.
X₂: Square feet of a house on the shore of Tawas Bay.
X₃: Number of bedrooms of a house on the shore of Tawas Bay.
The multiple regression model is Y= α + β₁X₁ + β₂X₂ + β₃X₃ + εi
a. Using software I've entered the raw data and estimated the regression coefficients:
^α= a= -5531.01
Represents the mean selling price of the houses when 0 bathrooms, 0 square feet and 0 bedrooms.
^β₁= b₁= -1386.21
Represents the modification of the mean selling price of the houses when the number of bathrooms increases in one unit and the square feet and number of bedrooms remain unchanged.
^β₂= b₂= 60.28
Represents the modification of the mean selling price of the houses when the square feet increase in one unit and the number of bathrooms and bedrooms remain unchanged.
^ β₃= b₃= 54797.08
Represents the modification of the mean selling price of the houses when the number of bedrooms increase in one unit and the number of bathrooms and square feet of the houses remain unchanged.
^Y= -5531.01 -1386.21X₁ + 60.28X₂ + 54797.08X₃
b)
R²= 0.55
R²Aj= 0.49
The coefficient of determination gives you an idea of how much of the variability of the dependent variable (Y) is due to the explanatory variables. Each time you add another explanatory variable to the regression the coefficient increases regarding of real contribution of the new variable. This could lead to thinking (wrongly) that the new variables are good to explain the dependent variable.
The adjusted coefficient of determination is a correction made to the raw coefficient of determination to have a more unbiased estimation of the effect the independent variables have over the dependent variable.
⇒ As you can see both coefficient are around 50%, which means that these explanatory variables
c)
The standard error estimate, this is the estimate of the population variance of the errors. In the ANOVA is represented by the Mean Square of the errors (MME)
Se²= MME= 3837640577.01
Se= 61948.6931
d) and f)
For the hypotheses tests for each slope the t- and p-values are:
α: 0.05
β₁: t= -0.06; p-value: 0.9528 ⇒ Do not reject H₀, the test is not significant.
β₂: t= 2.56; p-value: 0.0180 ⇒ Reject H₀, the test is significant.
β₃: t= 2.28; p-value: 0.0326 ⇒ Reject H₀, the test is significant.
e)
H₀: β₁= β₂= β₃
H₁: At least one βi is different from the others ∀ i=1, 2, 3
α: 0.05
F= 9.03
p-value: 0.0004
⇒ Reject H₀, the test is significant.
I hope it helps!