MATHEMATICS HIGH SCHOOL

Which of these equations represent functions? Check all that apply.A. y = (x - 2)^2 + 5
B. x^2 - 4y^2 = 1
C. y = 3x - 3
D. y = 4x^5

Answers

Answer 1
Answer: A is quadratic function
B is equation of a circle
C is equation of a line
D is hiperbolic funciotn

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(3 pt) Which shows the dimensions of two rectangular prisms that have volumes of 320 mm3 but different surface areas? A. 16 mm by 5 mm by 4 mm; 5 mm by 4 mm by 16 mm B. 10 mm by 4 mm by 6 mm; 6 mm by 10 mm by 4 mm C. 8 mm by 10 mm by 4 mm; 10 mm by 8 mm by 4 mm D. 8 mm by 10 mm by 4 mm; 20 mm by 8 mm by 2 mm

Answers

D. 8 mm by 10 mm by 4 mm
    20 mm by 8 mm by 2 mm

8 x 10 x 4 = 320 mm
20 x 8 x 2 = 320 mm

The area of a rectangular sign is 36 square meters. If the length of the sign is one meter more than twice the width, then what is the width of the sign?

Answers

l-length\nw-width\n\n2w=l-1\ \ \ \ \ /:2\n\nw=(l-1)/(2)\n\nArea:A=w* l;\ A=36\ m^2\n\nw* l=36\n\nput\ w=(l-1)/(2)\ into\ an\ equation\ w* l=36:\n\n(l-1)/(2)* l=36\ \ \ \ \ \ /*2

(l-1)* l=72\n\nl^2-l-72=0\n\nl^2-9l+8l-72=0\n\nl(l-9)+8(l-9)=0\n\n(l-9)(l+8)=0\iff l-9=0\ or\ l+8=0\n\nl=9\ or\ l=-8 < 0\n\nl=9\ m\ then\ w=(9-1)/(2)=(8)/(2)=4\ (m)\n\nAnswer:The\ width\ of\ the\ sign\ equal\ 4m.

Which expressions are equivalent to 16^x/4^x

Answers

16^1 = 4^2 16•1 = 4•4 16 = 16

What does the law of cosines reduce to when dealing with a right triangle

Answers

Answer:

Pythagoras Theorem

Step-by-step explanation:

For any triangle ABC, the law of cosine is given by

c^2=a^2+b^2-2ab\cos C

Now, let us suppose the triangle ABC is a right triangle having right angle at C.

Thus, C = 90°

Substituting this value in above formula

c^2=a^2+b^2-2ab\cos 90

We know that cos 90°= 0

Thus, the equation becomes

c^2=a^2+b^2-2ab\cdot0\n\nc^2=a^2+b^2

We can see that it reduces to Pythagoras Theorem.

Hence, we can conclude that law of cosines reduce to Pythagoras Theorem when dealing with a right triangle

3. We will now study probability distributions that can be obtained from the data. (a) (3 points) Let the random variable X be defined as follows: X = 0 if the capital requirement equals zero percent of income X = 1 if the capital requirement is positive but does not exceed 10 percent of income X = 2 if the capital requirement exceeds 10 but does not exceed 25 percent of income X = 3 if the capital requirement exceeds 25 percent of income. Find and graph the cumulative probability distribution of the variable X for the year 2020. (b) (2 points) Using the distribution from exercise 3(a), compute P(1 ≤ X < 3). (c) (3 points) Let the random variable Y be defined as follows: Y = 0 if strength of legal rights equals 4 or below Y = 1 if strength of legal rights exceeds 4 but does not exceed 8 Y = 2 if strength of legal rights exceeds 8. Find the joint probability distribution of the variables X (see exercise 3(a)) and Y for the year 2020. (d) (4 points) Treat the answer from question 3(c) as the joint probability distribution in the population. Using that distribution, what is the correlation between X and Y ?

Answers

(a)   1

(b)  0.3

(c)  0.15
(d)  0.27

(a) The cumulative probability distribution of the random variable X for the year 2020 is:
X = 0, P(X<=0) = 0.2
X = 1, P(X<=1) = 0.6
X = 2, P(X<=2) = 0.9
X = 3, P(X<=3) = 1
Graph:


(b) P(1 ≤ X < 3) = P(X<=2) - P(X<=1) = 0.9 - 0.6 = 0.3

(c) The joint probability distribution of the variables X and Y for the year 2020 is:
X = 0, Y = 0, P(X=0, Y=0) = 0.15
X = 0, Y = 1, P(X=0, Y=1) = 0.25
X = 0, Y = 2, P(X=0, Y=2) = 0.05
X = 1, Y = 0, P(X=1, Y=0) = 0.2
X = 1, Y = 1, P(X=1, Y=1) = 0.4
X = 1, Y = 2, P(X=1, Y=2) = 0.2
X = 2, Y = 0, P(X=2, Y=0) = 0.3
X = 2, Y = 1, P(X=2, Y=1) = 0.3
X = 2, Y = 2, P(X=2, Y=2) = 0.2
X = 3, Y = 0, P(X=3, Y=0) = 0.15
X = 3, Y = 1, P(X=3, Y=1) = 0.15
X = 3, Y = 2, P(X=3, Y=2) = 0.15

(d) Treating the answer from question 3(c) as the joint probability distribution in the population, the correlation between X and Y is 0.27.

  Learn more about probability

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Fifteen is __% of 120.

Answers

To do this,  divide 15 by 120.  You get 0.125  Then turn the decimal into a percent.  You do this by moving the decimal point to places to the RIGHT.  You get 12.5% as your answer.
It is 12.5%.

15 divided by 120 is 0.125, and in order to make it a percentage, multiply by 100% or move the decimal two places to the right.

Another way is to think about it and break it down. 60 is 50% of 120, and 15 is 25% of 60.
Since 15 is 25% of 60, which in turn in 50% of 120, 15 is 12.5%. 

I'm 1,000% sure it's correct.
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