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Debora [2.8K]
3 years ago
13

What type of graph is this?

Mathematics
1 answer:
murzikaleks [220]3 years ago
5 0

Answer:

The graph of a quadratic equation, or a parabola, looks like a U, an upside down U, a C, or a backwards C. We can use the following rules to determine what the graph of a given quadratic equation looks like. If y = ax2 + bx + c, and a is positive, then the graph of the equation is the shape of a U.

Step-by-step explanation:

bc

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect
Serggg [28]

Answer:

k = 1

P(x > 3y) = \frac{2}{3}

Step-by-step explanation:

Given

f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }  & { \text 0, { elsewhere. } } \end{array} \right.

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1

Where

{ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y  \leq x }

Substitute values for the interval of x and y respectively

So, we have:

\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1

Isolate k

k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1

Integrate y, leave x:

k \int\limits^2_{0} y {dx} \, [0,x/2]= 1

Substitute 0 and x/2 for y

k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1

k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1

Integrate x

k * \frac{x^2}{2*2} [0,2]= 1

k * \frac{x^2}{4} [0,2]= 1

Substitute 0 and 2 for x

k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1

k *[ \frac{4}{4} - \frac{0}{4} ]= 1

k *[ 1-0 ]= 1

k *[ 1]= 1

k = 1

Solving (b): P(x > 3y)

We have:

f(x,y) = k

Where k = 1

f(x,y) = 1

To find P(x > 3y), we use:

\int\limits^a_b \int\limits^a_b {f(x,y)}

So, we have:

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy

P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0  dxdy

Integrate x leave y

P(x > 3y) = \int\limits^2_0  x [0,y/3]dy

Substitute 0 and y/3 for x

P(x > 3y) = \int\limits^2_0  [y/3 - 0]dy

P(x > 3y) = \int\limits^2_0  y/3\ dy

Integrate

P(x > 3y) = \frac{y^2}{2*3} [0,2]

P(x > 3y) = \frac{y^2}{6} [0,2]\\

Substitute 0 and 2 for y

P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}

P(x > 3y) = \frac{4}{6} -\frac{0}{6}

P(x > 3y) = \frac{4}{6}

P(x > 3y) = \frac{2}{3}

8 0
3 years ago
Which of the following are the correct answers??
Amanda [17]
Either B and D. Both seem pretty random lol
6 0
3 years ago
Read 2 more answers
Solve! Image attached! Geometry, dumb/incorrect answers will get account, and answer reported, please answer all the way through
neonofarm [45]

Answer:

+) m∠1 = m∠2 (vertical angles theorem)

+) m∠2 and ∠3 are supplementary (because a//b)

+)m∠3 = m∠4 (alternate interior angles theorem, c//d)

=> ∠1 and ∠4 are supplementary (substitution property of equality)

Step-by-step explanation:

5 0
3 years ago
More than half of the students were excited about the project. In fact, were excited. If 10 students were not
alina1380 [7]

Answer:1/2 divided by 10

Step-by-step explanation:

8 0
3 years ago
Simplify:2 1/3(4 1/4÷2)
kondor19780726 [428]

Answer:

4\frac{23}{24}

Step-by-step explanation:

We want to simplify:

2\frac{1}{3}(4\frac{1}{4}\div 2)

We convert to improper fraction to get:

\frac{7}{3}(\frac{17}{4}\div 2)

We now deal with the parenthesis by multiplying with the reciprocal of 2.

\frac{7}{3}(\frac{17}{4}\times \frac{1}{2} )

This gives:

\frac{7}{3}(\frac{17}{8})

We multiply to get:

\frac{7}{3}(\frac{17}{8})

This simplfies to:

\frac{119}{24}

We write as mixed number

4\frac{23}{24}

4 0
3 years ago
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