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Blizzard [7]
3 years ago
8

Can you help me? Please show how you work it out

Mathematics
1 answer:
qaws [65]3 years ago
8 0

Swimmers: 50 children ÷ 12 = 4.2  (you can't have a decimal so you must round up)

5 adults are need.

Dancers: 32 children ÷ 15 = 2.1  (you can't have a decimal so you must round up)

3 adults are need.

Minimum number of adults needed is: 5 for swimmers + 3 for dancers = 8 adults

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andriy [413]

Answer:

R-squared is a statistical measure of how close the data are to the fitted regression line.

Step-by-step explanation:

It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. ... 100% indicates that the model explains all the variability of the response data around its mean.

4 0
3 years ago
4.6.3 Test (CST): Linear Equations
inn [45]

The slope of the green line if the lines are perpendicular is -1/4

<h3>Perpendicular lines</h3>

For two lines two be perpendicular, the product of their slope must be -1. Let the slope of the red and green line be m1 and m2.

Given the following

Slope of red line = 4

According the definition

4m2 = -1

m2 = -1/4

Hence the slope of the green line if the lines are perpendicular is -1/4

Learn more on perpendicular lines here: brainly.com/question/1202004

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5 0
2 years ago
Algebra 1 help!!!!!!!!!
baherus [9]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of expo functions.

f(x) = -8(2)^x - 12 ,

for f(0) ,here substitute x = 0

so we get as ,

==> f(0) = -8(2)^0 -12

==> f(0) = -8-12

==> f(0) = -20

hence, f(0) = -20

6 0
3 years ago
Read 2 more answers
Simplify.<br> 4/x^2-35 - 1/x+5
Oksana_A [137]
If I understand your question well, I suppose you mean \frac{4}{ x^{2} -35} -  \frac{1}{x+5}
\frac{4}{ x^{2} -35} - \frac{1}{x+5} =  \frac{4}{(x+5)(x-7)}- \frac{1}{x+5} \\ = \frac{1}{x+5}( \frac{4}{x-7}-1) \\ = \frac{1}{x+5}( \frac{4}{x-7}- \frac{x-7}{x-7}) \\ = \frac{1}{x+5}( \frac{4-x+7}{x-7}) \\ = \frac{1}{x+5}( \frac{11-x}{x-7}) \\ = \frac{11-x}{ x^{2} -35}
6 0
3 years ago
Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).
alexira [117]

Let f(x,y,z)=z-\ln(x+y). The gradient of f at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

\nabla f(1,0,0)\cdot(x-1,y,z)=0

Compute the gradient:

\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)

Evaluate the gradient at the given point:

\nabla f(1,0,0)=(-1,-1,1)

Then the equation of the tangent plane is

(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}

7 0
4 years ago
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