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3 years ago

What is 34 + 35? does anyone like anime lol.

Mathematics

2 answers:

jeka57 [31]3 years ago

5 0

There is no carrying so you add 5 and 4 with is 9 and 3 and 3 which 6 so the answer is 69

AVprozaik [17]3 years ago

3 0

Answer:

I like anime:)

Step-by-step explanation:

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Suppose in this semester, our Exam 1 average was about 86 with an SD of about 10. Suppose the correlation between our Exam 1 and

Oduvanchick [21]

Answer:

the slope of the regression equation for predicting our Exam 2 scores from Exam 1 scores is 0.492

And the y-intercept of the regression equation for predicting our Exam 2 scores from Exam 1 is 33.688

Step-by-step explanation:

Given the data in the question;

mean X" = 86

SD σx = 10

Y" = 76

SD σy = 8.2

r = 0.6

Here, Exam 2 is dependent and Exam 1 is independent.

The Regression equation is

y - Y" = r × σy/σx ( x - x" )

we substitute

y - 76 = 0.6 × 8.2/10 ( x - 86 )

y - 76 = 0.492( x - 86 )

y - 76 = 0.492x - 42.312

y = 0.492x - 42.312 + 76

y = 0.492x + 33.688

Hence, the slope of the regression equation for predicting our Exam 2 scores from Exam 1 scores is 0.492

And the y-intercept of the regression equation for predicting our Exam 2 scores from Exam 1 is 33.688

4 0

3 years ago

Square root of a number minus five.

grandymaker [24]

Answer:

it Doesnt have a square root

Step-by-step explanation:

7 0

3 years ago

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240<br><br> dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

3 years ago

Pls help!!!!<br> Find the center, vertices, foci, and asymptotes of the hyperbola. 9y^2-36x^2=324

ICE Princess25 [194]

Answer:

Center: (0, 0)

Vertices: (0, 6), (0, -6)

Foci: (0, 3√5), (0, -3√5)

Asymptotes: y = 2x, y = -2x

Step-by-step explanation:

7 0

2 years ago

Y - 4 = 3/4(x+8) in standard form

Nesterboy [21]

To write a linear equation in standard form, move each variable term to the left side of the equation and simplify. Ax +By = C

Since x is on the right side of the equation, switch the sides so it is on the left side of the equation. 3/4 * (x+8) = y-4

Move all terms containing variables to the left side of the equation

3x/4 +6 = y-4

Move the y to the left side of the equation

3x/4+6-y=-4

Move all terms not containing a varbile to the right side

Move the 6 to the right side

3x/4-y=-6-4

Subtract 4 from -6

3x/4-y=-10

Find the LCD of A,B,C and you get 4

Then you * 4 by everything

3x/4*4-y*4=-10*4

Simplify each term

3x-4y=-10*4

Multiply -10 by 4

3x-4y=-40

8 0

3 years ago

Read 2 more answers