Answer: y=-x+-8
Step-by-step explanation:
Proof is in the screenshot below
Answer:option B & D are correct
Step-by-step explanation:
Given options are
A B C D E
Tether ball pole shadow 1.35 1.8 3.75 0.6 2
Flagpole shadow 3.4 4.08 8.35 1.36 4.8
Ratio of corresponding height & shadows should be same
3 meter tall tether ball pole and a 6.8 tall flagpole
=> Ratio = 6.8/3 = 68/30 = 34/15
A - 3.4/1.35 = 340/135 ≠ 34/15( 306/135)
B - 4.08/1.8 = 408/180 = 34/15 = 34/15 ( Correct option)
C 8.35/3.75 = 835/375 = 167/75 ≠ 34/15( 170/75)
D 1.36/0.6 = 136/60 = 34/15 = 34/15 ( Correct option)
E 4.8/2 = 48/20 = 12/5 = 36/15 ≠ 34/15
option B & D are correct
49x^2 + 64y^2 = 3136 . . . (1)
y - 6 = (1/4)x^2 . . . (2)
From (2), x^2 = 4(y - 6) = 4y - 24 . . . (3)
Putting (3) into (1) gives
49(4y - 24) + 64y^2 = 3136
196y - 1176 + 64y^2 = 3136
64y^2 + 196y - 4312 = 0
y = 6.819
x = sqrt(4(6.819) - 24)
x = 1.81
The possible point of collision is (1.81, 6.819)
Answer:
(1) 60
(2) 487635
Step-by-step explanation:
(1) Distinguishable shuffles means that all the cards must be different from each other. We have 9 different cards types having 4 cards each and 1 card having 24 copies of it. So, in a distinguishable shuffle, we can get 1 card each from the 9 different cards (C1 - C9) and another card from the set of 24 identical cards. So we will have a total of 10 cards.
The possible ways of getting these cards are:
9 x (⁴C₁) + ²⁴C₁
(9 x 4) + 24 = 60
There are 60 distinguishable shuffles.
(2) Here we are simply asked to calculate the number of 4-card combinations that can be drawn from the deck of 60 cards. It is not mentioned that the cards must be distinguishable so,
Possible hands of four cards = ⁶⁰C₄ = 487635
Answer: Option D : All of the above.
Step-by-step explanation:
(1) The Cumulative Link Models assumptions can be summarized as, conditional on (x1, … xk)
(2) Whether normality of error u, and thus normality of y conditional on (x1, … xk) can be assumed, is an empirical matter
and (3) Normality of error u translates into normal sampling distributions of the OLS estimators are all correct statements describing linear regression.
