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sashaice [31]
2 years ago
11

Please I need help fast!!!

Mathematics
1 answer:
never [62]2 years ago
5 0

The answer is c. because all of the other answers arne tpossible due to where they graphed it

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Which expression is rational?
horrorfan [7]

Answer:

square root of 49

Step-by-step explanation:

square root of 2 is 1.41421356237

square root of 14 is 3.74165738677

square root of 49 is 7

square foot of 49 is the correct one because 6.27316543222718… keeps going and does not stop the other 2 is irrational and square root of 49 is 7 basically meaning is rational

Hope this helped! :)

May i have brainliest!

5 0
3 years ago
Adriano loves anime and collects toys from his favorite series. each month he gains 4 toys. if adriano has 3 toys when he starts
skad [1K]
The equation for this would be 4x+3=y where x is months and y is your amount of toys after x months.
plug in 4 for x -> 4(4)+3=y
solve -> 16+3=y
y=19
8 0
3 years ago
Read 2 more answers
Please guys I keep getting this question wrong What letter is located at approximately right answers only
liubo4ka [24]
Answer:
B) F
You put √22 in a calculator and it gives you ≈ 4.69
4 0
2 years ago
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
√(4²+7²) PLS HELP WILL GIVE BRAINLIEST FOR QUICKEST AND CORRECT ANSWER!!!!
cupoosta [38]

Answer:

8.06

Step-by-step explanation:

Hope it helps

7 0
3 years ago
Read 2 more answers
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