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3 years ago

Which statements are true about triangle ABC and its translated image, A'B'C'? Check all that apply.

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

4 0

This is the concept of translation;
Given that triangle A (-1,3), B(-1,1), C(3,1) has been translated to A'(2,-1), B'(2,-4), C(6,-4)
the from the diagram we see that the coordinates of the triangle ABC have been moved by 3 units to the right while the y coordinates have been moved by 5 units down to form A',B',C'.
We therefore conclude that the correct statement is:
Triangle ABC has been translated 3 units to the right and 5 units down.

adelina 88 [10]3 years ago

3 0

Answer:

The rule for the translation can be written as T3, –5(x, y).

Triangle ABC has been translated 3 units to the right and 5 units down.

Step-by-step explanation:

Just took the test.

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The two odd numbers are 15 and 17

Step-by-step explanation:

Given

Let the odd numbers be represented with x and y

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$x + 5y = 92$

Required

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Since x and y are consecutive odd numbers and x is greater, then

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The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2 respectively. Let A de

antoniya [11.8K]

Answer:

a) $P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b) $P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c) $P(A') = 1-P(A) =1-0.4=0.6$

d) $P(A \cup B) =0.4 +0.8-0.2 =1.0$

e) The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

Step-by-step explanation:

We have the following space provided:

$S= [a,b,c,d,e]$

With the following probabilities:

$P(a) =0.1, P(b)=0.1, P(c) =0.2, P(d)=0.4, P(e)=0.2$

And we define the following events:

A= [a,b,c], B=[c,d,e]

For this case we can find the individual probabilities for A and B like this:

$P(A) = 0.1+0.1+0.2 = 0.4$

$P(B) =0.2+0.4+0.2=0.8$

Determine:

a. P(A)

$P(A) =P(a)+P(b) +P(c)= 0.1+0.1+0.2 = 0.4$

b. P(B)

$P(B) =P(c) +P(d)+P(e)=0.2+0.4+0.2=0.8$

c. P(A’)

From definition of complement we have this:

$P(A') = 1-P(A) =1-0.4=0.6$

d. P(AUB)

Using the total law of probability we got:

$P(A \cup B) =P(A) +P(B)-P(A \cap B)$

For this case $P(A \cap B) = P(c) =0.2$ , so if we replace we got:

$P(A \cup B) =0.4 +0.8-0.2 =1.0$

e. P(AnB)

The intersection between the set A and B is the element c so then we have this:

$P(A \cap B) = P(c) =0.2$

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