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Lena [83]
3 years ago
11

Quadrilateral ABCD is transformed according to the rule (x, y) → (y, –x). Which is another way to state the transformation?

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
5 0
The answer is RO 270
scoray [572]3 years ago
4 0
The answer
there is no more explanation about this question, the solution can be found easily by applying the rule of transformation of rotation

for example, the properties of rotation transformation are:
A rotation preserves length but does not necessarily preserveslope of a line.
A 90° rotation ( 1/4 turn) anticlockwise about the origin changesthe point (x; y) to (-y; x).
A 180° rotation ( 1/2 turn) clockwise or anticlockwise about theorigin changes the point (x; y) to (-x;-y).
A 270° rotation ( 3/4 turn) anticlockwise changes about the originthe point (x; y) to (y;-x).
 so the answer is 
<span>R0, 270°</span>
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Answer:

0.2

Step-by-step explanation:

20/100=0.2


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A researcher plants 22 seedlings. After one month, independent of the other seedlings, each seedling has a probability of 0.08 o
Andrews [41]

Answer:

E(X₁)= 1.76

E(X₂)= 4.18

E(X₃)= 9.24

E(X₄)= 6.82

a. P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= 0.00022

b. P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= 0.000001

c. P(X₁≤2) = 0.7442

Step-by-step explanation:

Hello!

So that you can easily resolve this problem first determine your experiment and it's variables. In this case, you have 22 seedlings (n) planted and observe what happens with the after one month, each seedling independent of the others and has each leads to success for exactly one of four categories with a fixed success probability per category. This is a multinomial experiment so I'll separate them in 4 different variables with the corresponding probability of success for each one of them:

X₁: "The seedling is dead" p₁: 0.08

X₂: "The seedling exhibits slow growth" p₂: 0.19

X₃: "The seedling exhibits medium growth" p₃: 0.42

X₄: "The seedling exhibits strong growth" p₄:0.31

To calculate the expected number for each category (k) you need to use the formula:

E(XE(X_{k}) = n_{k} * p_{k}

So

E(X₁)= n*p₁ = 22*0.08 = 1.76

E(X₂)= n*p₂ = 22*0.19 = 4.18

E(X₃)= n*p₃ = 22*0.42 = 9.24

E(X₄)= n*p₄ = 22*0.31 = 6.82

Next, to calculate each probability you just use the corresponding probability of success of each category:

Formula: P(X₁, X₂,..., Xk) = \frac{n!}{X_{1}!X_{2}!...X_{k}!} * p_{1}^{X_{1}} * p_{2}^{X_{2}} *.....*p_{k}^{X_{k}}

a.

P(X₁=3, X₂=4, X₃=6;0.08,0.19,0.42)= \frac{22!}{3!4!6!} * 0.08^{3} * 0.19^{4} * 0.42^{6}\\ = 0.00022

b.

P(X₁=5, X₂=5, X₄=7;0.08,0.19,0.31)= \frac{22!}{5!5!7!} * 0.08^{5} * 0.19^{5} * 0.31^{7}\\ = 0.000001

c.

P(X₁≤2) = \frac{22!}{0!} * 0.08^{0} * (0.92)^{22} + \frac{22!}{1!} * 0.08^{1} * (0.92)^{21} + \frac{22!}{2!} * 0.08^{2} * (0.92)^{20} = 0.7442

I hope you have a SUPER day!

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sukhopar [10]

Answer:

60 mph

Step-by-step explanation:

Given;

Total distance covered d = 440 miles

Average speed in the first 4 hours v1 = 50 mph

Total time taken for the whole trip t = 8:00 to 4:00pm

t = 16:00-8:00 = 8 hours

Firstly we need to calculate the distance covered during the first 4 hours;

d1 = average speed × time = v1 × t1

t1 = 4

d1 = 50 × 4 = 200 miles

Then we need to calculate the distance covered in the second period of the journey;

d = d1 + d2

d2 = d - d1

Substituting the values;

d2 = 440 - 200

d2 = 240 miles

The time taken for the second period of travel t2;

t2 = t -t1 = 8-4

t2 = 4 hours

Average speed = distance travelled ÷ time taken

The average speed for the second part of the trip v2 is;

v2 = d2 ÷ t2

Substituting t2 and d2;

v2 = 240 miles ÷ 4 hours

v2 = 60 mph

The average speed during the second part of the trip is 60 mph

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Paladinen [302]
2 + 1 = 3
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Solution:

<u>In this case, we are given:</u>

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Looking at Option A, we can tell that its <u>slope is -2 and y-intercept is 0.</u>

Since this is matching with the info we are given, Option A is correct.

7 0
2 years ago
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