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ss7ja [257]
3 years ago
10

Can someone anyone help me??? This is an example problem for how my teacher wants our work to be shown with this layout I don’t

know what to do please someone help me

Mathematics
1 answer:
Lana71 [14]3 years ago
7 0
a =  \frac{1}{4}
b = -6
c = -5
d = 1

I think that this is what your teacher is trying to ask. 
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2x exponent of 2 divided by 27 x=18
Paha777 [63]

Answer:

24

Step-by-step explanation:

3 0
3 years ago
Find the equation of the line in gradient form that passes through the points (3,0) and (-2,4)
Aleksandr-060686 [28]

Answer:

y = (-4/5)x + 12/5

Step-by-step explanation:

Slope = (4-0)/(-2-3) = 4/-5 = -4/5

y = (-4/5)x + c

When x = 3, y = 0

0 = (-4/5)(3) + c

c = 12/5

y = (-4/5)x + 12/5

5y = -x + 12

5y + x = 12

4 0
3 years ago
Each morning, Hungry Harry eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4,
mamaluj [8]

Answer:

Mean = 35

Variance = 291.7

Step-by-step explanation:

Data provided in the question:

X : 1, 2, 3, 4, 5, 6

All the data are independent

Thus,

The mean for this case will be given as:

Mean, E[X] = \frac{\textup{Sum of all the observations}}{\textup{Total number of observations}}

or

 E[X] = \frac{\textup{1+2+3+4+5+6}}{\textup{6}}

or

E[X] = 3.5

For 10 days, Mean = 3.5 × 10 = 35

And,

variance = E[X²] - ( E[X] )²

Now, for this case of independent value,

E[X²] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{\textup{6}}

or

E[X²] = \frac{1+4+9+16+25+36}{\textup{6}}

or

E[X²] = \frac{91}{\textup{6}}

or

E[X²] = 15.167

Therefore,

variance = E[X²] - ( E[X] )²

or

variance = 15.167 - 3.5²

or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

= 291.7

3 0
3 years ago
How to solve 4k-1.5k=50
padilas [110]
Combine like terms-- you can combine the k's:
4k-1.5k=2.5k

Now you have:
2.5k=50

Simply divide both sides by 2.5 to find the value of k:
\frac{2.5k}{2.5}=\frac{50}{2.5}\\\\k=20
6 0
3 years ago
I need help indicating the exterior angle
levacccp [35]

Check the picture below.

7 0
3 years ago
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