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2 years ago

Triangle ABC was dilated and translated to form similar triangle A'B'C'.

Mathematics

2 answers:

Sergeu [11.5K]2 years ago

7 0

correct answer is 5/2

Hunter-Best [27]2 years ago

3 0

Dilation refers to a non rigid motion where a figure is transform and its image has the same form but a different size measure.

On this exercise is asked to find the scale factor by which the triangle ABC was
dilated to produce the triangle A'B'C'.
Dilation is define by the rule (x,y)-- (kx, ky) where k represents the scale factor.

As can be see on the picture the dilation produce was an enlargement meaning that the image is larger that the preimage.
Of this form you can discard the choices A and B as possible solutions.

Lets try 5/2 as the possible scale factor:
(x,y)-- (kx, ky)
A(0,2)--(5/2(0),5/2(2))=A'(0,5)
B(2,2)--(5/2(2),5/2(2))=B'(5,5)
C(2,0)--(5/2(2),5/2(0))=C'(5,0)

Lets try 5/1 or 5 as the scale factor:
A(0,2)--(5(0),5(2))=A'(0,10)
B(2,2)--(5(2),5(2))=B'(10,10)
C(2,0)--(5(2),5(0))=C'(10,0)

As said at the beginning of the question the triangle was not only dilated.
After a dilation and a translation, the scale factor of the dilation is letter C or 5/2.

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Answer:

99% Confidence interval: (0.185,0.375)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 150

Number of cars that have faulty catalytic converters, x = 42

$\hat{p} = \dfrac{x}{n} = \dfrac{42}{150} = 0.28$

99% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Putting the values, we get:

$0.28\pm 2.58(\sqrt{\frac{0.28(1-0.28)}{150}}) = 0.28\pm 0.095\\\\=(0.185,0.375)$

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2 years ago

Ken and Hamid run around a track. It takes Ken 80 seconds to complete a lap. It takes Hamid 60 seconds to complete a lap. Ken an

likoan [24]

The <em><u>correct answer</u></em> is:

Ken will have run 3 laps and Hamid will have run 4.

Explanation:

To find this, we first find the number of seconds that will have passed when they meet again. We use the LCM, or least common multiple, for this. First we find the prime factorization of each number:

80 = 10(8)

10 = 5(2)

8 = 2(4)

4 = 2(2)

80 = 2(2)(2)(5)(2)

60 = 10(6)

10 = 5(2)

6 = 2(3)

60 = 2(2)(3)(5)

For the LCM, we multiply the common factors by the uncommon. Between the two numbers, the common factors are 2, 2 and 5. This makes the uncommon 2, 2, and 3, and makes our LCM

2(2)(5)(2)(2)(3) = 240

This means every 240 seconds they will both be at the start line.

Since Ken completes a lap in 80 seconds, he completes 240/80 = 3 laps in 240 seconds.

Since Hamid completes a lap in 60 seconds, he completes 240/60 = 4 laps in 240 seconds.

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2 years ago

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Step-by-step explanation:

Given that,

A piece of buttered toast falls to the floor 17 times. The toast landed buttered side up 6 times.

It means that the total number of outcomes are 17

We need to find the probability that the toast lands buttered side down. Favourable oucome is 17-6 = 11

So, probability is given by :

$P(E)=\dfrac{\text{favourable outcomes}}{\text{total no of outcomes}}$

$P(E)=\dfrac{11}{17}$

So, the probability that the toast lands buttered side down is 11/17.

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2 years ago

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$

Therefore, Probability of atleast one of 12 student has food allergies

= $1-(\frac{93}{100})^{12}$

= $0.581403702521$

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