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maw [93]
3 years ago
7

Directions: Factor 15x2 – 25x - 40

Mathematics
1 answer:
Delvig [45]3 years ago
7 0
5(x + 1) (3x - 8)

Hope this helps!
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Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)
irina1246 [14]

Answer:

Maximize C =9x + 7y

8x + 10y \leq 17

11x + 12y\leq 25

and x ≥ 0, y ≥ 0

Plot the lines on graph

8x + 10y \leq 17

11x + 12y\leq 25

x\geq 0

y\geq 0

So, boundary points of feasible region are (0,1.7) , (2.125,0) and (0,0)

Substitute the points in  Maximize C

At  (0,1.7)

Maximize C =9(0) + 7(1.7)

Maximize C =11.9

At  (2.125,0)

Maximize C =9(2.125) + 7(0)

Maximize C =19.125

At   (0,0)

Maximize C =9(0) + 7(0)

Maximize C =0

So, Maximum value is attained at   (2.125,0)

So, the optimal value of x is 2.125

The optimal value of y is 0

The maximum value of the objective function is 19.125

4 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
The 6 consecutive integers below add up to 447. x − 2 x − 1 x x + 1 x + 2 x + 3 what is the value of x ? f. 72 g. 73 h. 74 j. 75
IrinaVladis [17]
(x-2) + (x-1) + x + (x+1) + (x-2) + (x+3) = 447
6x + 3 = 447
6x = 447 - 3
6x = 444
x = 444/6
x = 74
3 0
3 years ago
5 cows went for a walk what did they get?<br> 2343=C<br> 23432=D
Fynjy0 [20]

Answer:

i dont know

Step-by-step explanation:

they got nick popkes

7 0
2 years ago
Solve : 3[3[3]-2] square 2 -3[6-4] square 2<br> please help me
Hoochie [10]
What are the brackets for? o and im srry for asking but could u mark this "answer" brainliest?? i just need one more to lvl up Im srry ik this isnt an answer but it would mean the world to me if u did
:D
5 0
3 years ago
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