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kipiarov [429]
3 years ago
5

ΔTOY has coordinates T (−3, 4), O (−4, 1), and Y (−2, 3). A translation maps point T to T' (−1, 1). Find the coordinates of O' a

nd Y' under this translation.
O' (−2, −2); Y' (0, 0)
O' (−1, −1); Y' (1, 1)
O' (0, 0); Y' (−2, −2)
O' (1, 1); Y' (−1, 0)
Mathematics
2 answers:
polet [3.4K]3 years ago
5 0

Answer:  The correct option is (A) O' (−2, −2); Y' (0, 0).

Step-by-step explanation:  Given that the co-ordinates of the vertices of ΔTOY are T(−3, 4), O (−4, 1), and Y (−2, 3). A translation maps point T to T' (−1, 1).

We are to find the co-ordinates of the points O' and Y'.

The given transformation from T to T' is

T(−3, 4)   ⇒   T' (−1, 1).

Let,  (−3 + x, 4 + y) =  (-1, 1).

So,

-3+x=-1\\\\\Rightarrow x=2

and

4+y=1\\\\\Rightarrow y=-3.

That is, the transformation rule is

(a, b) ⇒ (a+2, b-3).

Therefore,

co-ordinates of O' are (-4+2, 1-3) = (-2, -2),

and

co-ordinates of Y' are (-2+2, 3-3) = (0, 0).

Thus, the required co-ordinates of O' and Y' are (-2, -2) and (0, 0) respectively.

Option (A) is correct.

Setler79 [48]3 years ago
3 0
Given:
T (-3,4) ; O (-4,1) ; Y (-2,3)
T'(-1,1) ; 

T
-3 move forward 2 points to reach -1
4 move forward 3 points to reach 1

O
-4 move forward 2 points to reach -2
1 move forward 3 points to reach -2

Y
-2 move forward 2 points to reach 0
3 move forward 3 points to reach 0.

O'(-2,-2) ; Y'(0,0)   1st option.

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Answer:

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Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                               NyQuil          Robitussin       Triaminic     Total

No relief                    11                     13                      9               33

Some relief               32                   28                     27              87

Total relief                7                       9                      14               30

Total                         50                    50                    50              150

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the three remedies

H1: There is a difference in the three remedies

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*33}{150}=11

E_{2} =\frac{50*33}{150}=11

E_{3} =\frac{50*33}{150}=11

E_{4} =\frac{50*87}{150}=29

E_{5} =\frac{50*87}{150}=29

E_{6} =\frac{50*87}{150}=29

E_{7} =\frac{50*30}{150}=10

E_{8} =\frac{50*30}{150}=10

E_{9} =\frac{50*30}{150}=10

And the expected values are given by:

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No relief                    11                     11                       11               33

Some relief               29                   29                     29              87

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Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(3-1)=4

And we can calculate the p value given by:

p_v = P(\chi^2_{4,0.05} >3.81)=0.43233

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(3.81,4,TRUE)"

Since the p values is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences between the 3 remedies analyzed.

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VMariaS [17]
The difference is 181 degrees.

Subtract 146 by -35 (146 - -35)
When a negative it being subtracted replace it with a positive because two negatives make a positive (146+35) which gets you 181
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