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3 years ago

Which of the following steps were applied to abcd to obtain abcd

Mathematics

2 answers:

bulgar [2K]3 years ago

7 0

Answer:

Step-by-step explanation:

Move to the right 2 units,

Move down 2 units.

Damm [24]3 years ago

7 0

Answer:

Option (c) is correct.

Shift 2 units right and 3 units down.

Step-by-step explanation:

Given : A graph showing two quadrilaterals ABCD and A'B'C'D'

We have to choose the correct option from the given options that describes the steps applied to ABCD to obtain A'B'C'D'.

Translation is moving every coordinate of the image in same direction and in same distance.

Here, Coordinate of A is (3,5) and Coordinate of A' is (5,2)

Thus, To move A to A' we have shifted x coordinate 2 units to right and y coordinate 3 units down.

Similarly, we can check for each point

Consider Coordinate of B is (4,8) and Coordinate of B' is (6,5)

Thus, To move B to B' we have shifted x coordinate 2 units to right and y coordinate 3 units down.

Thus, The steps applied to ABCD to obtain A'B'C'D' is shifted x coordinate 2 units to right and y coordinate 3 units down.

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I need this quickly please

IRINA_888 [86]

Answer:

y = 15x + 20 or 15x - y + 20 = 0

x = 11

Step-by-step explanation:

Line is passing through the points (0, 20) & (2, 50)

Slope of line = (50 - 20)/(2 - 0) = 30/2 = 15

Equation of line

y - 20 = 15(x - 0)

y - 20 = 15x

y = 15x + 20 or 15x - y + 20 = 0

Plug y = 185 in above equation, we find:

185 = 15x + 20

185 - 20 = 15x

165 = 15x

x = 165/15

x = 11

5 0

3 years ago

Simplify.

spin [16.1K]

B. 1.3

Hope that helped:)

5 0

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

3 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$

5 0

4 years ago

The slope please. it must be one of the four options.

Mazyrski [523]

The answer to the question is a

3 0

4 years ago