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Advocard [28]
3 years ago
12

polygon P'Q'R'S'T' shown on the grid below is an image of polygon PQRST after dilation with a scale factor of 3, keeping the ori

gin as the center of dilation: ​

Mathematics
1 answer:
ruslelena [56]3 years ago
6 0

Answer:  d) SR and S'R' have the ratio 1:3

<u>Step-by-step explanation:</u>

In order for the polygons to be similar, they must have congruent angles and proportional side lengths.

a) ∠Q and ∠Q' have the ratio 1:3          

  FALSE - The angles must be congruent (not proportional)

b) TS and T'S' have equal lengths

 FALSE - We can see that there is a dilation so they cannot be congruent.

c) RT and R'T' have equal lengths

 FALSE - We can see that there is a dilation so they cannot be congruent.

d) SR and S'R' have a ratio of 1:3

 TRUE! - The sides are proportional so we can use this to prove similarity.

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All your doing here is 37-19 adding negative is the same has munising 

So 37-19=18
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2 years ago
The smallest square number which can be divisible by 11 and 5​
Elena L [17]

Answer:

3025

Step-by-step explanation:

let's look at it in this concept.

For a number to be divisible by 11 and 5. it must be a multiple of the LCM of 11 and 5.

LCM of 11 and 5=55

therefore the number is 55x, where x is a positive integer.

it is a said that the number is a perfect square

therefore the square root of 55x must be an integer.

\sqrt{55x}  =  \sqrt{5 \times 11 \times x}

the smallest value of x to make 55x a perfect square is....

x = 11 \times 5 = 55

Therefore the number is.... .

55x = 55(55) = 3025

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6 0
3 years ago
The measure of A is 75°, and the measure of B is 105°. What is the relationship of angles A and B?
Leni [432]

Answer:

D supplementary angles

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Surveys, observational studies, and experiments are examples of
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Answer: B data gathering

Step-by-step explanation:

all of those examples are examples of gathering data

7 0
3 years ago
Two litters of a particular rodent species have been born, one with two brownhaired and one gray-haired (litter 1), and the othe
anyanavicka [17]

Answer:

a. So probability that the animal chosen is brown-haired = 0.633

b. Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A) = 0.5263.

Step-by-step explanation:

Being that We are given,event of Brown hair with two disjoint events, one is { ( BrownHair ) ∩ ( Litter 1) } and the other is { ( BrownHair ) ∩ ( Litter 2) } .

a) To find the probability that the animal chosen is brown-haired,

Let A : we choose a brown-haired rodent and B : we choose litter1.

So using the axioms of probability, we can write

P(A) = P(A | B) * P(B) + P(A | Bc) * P(Bc)

Making use of the given information, we get;

number of brown haired rodents in litter 2 P(AB) Total number of rodents in litterl

and

P(A |B^{c}) =\frac{\text{number of brown haired rodents in litter2}}{\text{Total number of rodents in litter2}}= \frac{3}{5}

And also it is given that we choose litter at random ,so P(B) = P(Bc ) = 1/2

So now we plug these values in the equation of P(A) and get

P(A) = (\frac{2}{3}*\frac{1}{2}) + (\frac{3}{5}*\frac{1}{2}) = \frac{2}{6}+\frac{3}{10} = 0.633

So probability that the animal chosen is brown-haired = 0.633

b) Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A)

Lets make use of Bayes rule to find this conditional probability,

So using theorem we get,

P(B|A) = \frac{P(A|B)*P(B)}{P(A|B)*P(B)+P(A|B^{c})*P(B^{c})}

P(B|A) = \frac{(1/2)*(2/3)}{[(1/2)*(2/3)]+[(1/2)*(3/5)]} = \frac{10}{19} = 0.5263

Thus, Given that a brown-haired offspring was selected, probability that the sampling was from litter1 = P(B|A) = 0.5263.

5 0
3 years ago
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