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Alik [6]
3 years ago
10

Solving slope (1,1/2),(3,2)

Mathematics
1 answer:
Lina20 [59]3 years ago
3 0
2-1/2       3/2       3
-------   =  -----  =  ---
3-1            2         4
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55% of the 880 students at julian's school are male. How many male students are in julian's school?
Oliga [24]

Answer:

448

Step-by-step explanation:

55 Divided by 880 is 448

4 0
3 years ago
For a given input value v, the function f outputs a value u to satisfy the following equation.
tigry1 [53]

Answer: -4v+9

Step-by-step explanation:

u - 5 = -4(v - 1)

u - 5 = -4v + 4

+5 +5

u = -4v + 9

8 0
3 years ago
For your fundraiser, you decide to hold a bake sale. In order to make bakery items, you had to purchase 12 pounds of sugar and 1
madam [21]
The question seems to be pointing towards a multiple choice response but I will do my best. The first equation would be 15x+12y=9.30 and the second would be 10x+4y=4.60, x represents flour and y represents sugar.

Hope this helps.
3 0
3 years ago
HELP PLEASE ASAP<br><br><br> Solve for x in the diagram below
Leni [432]

Answer:

29

Step-by-step explanation:

(x+12) + 100 + x = 180

           2x + 112 = 180

Subtract 112 from both sides

                    2x = 58

Divide 2 from both sides

                      x = 29

3 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
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