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qaws [65]
2 years ago
9

Please help me solve this

Mathematics
2 answers:
Mrrafil [7]2 years ago
6 0

Answer:

i was so bad a geometry that i cried everyday just so you know

Step-by-step explanation:

NNADVOKAT [17]2 years ago
3 0

Answer:

:0:0:0:0:0(*_*)(*_*)(*_*)(*_*)

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Which ordered pairs in the form (x, y) are solutions to the equation 
Artemon [7]
A and c are the correct answers, you just have to substitute the options in and see which one equals 10
4 0
3 years ago
Read 2 more answers
Find the sum. 2/5(d-10)-2/3(d+6)
alukav5142 [94]

Answer:

-4d

Step-by-step explanation:

L.c.m=15

15×2/5(d-10)-2/3×15(d+6)

3×2(d-10)-2×5(d+6)

6(d-10)-10(d+6)

6d-60-10d-60

6d-10d-60-60

-4d-0

-4d

5 0
2 years ago
Solve for k: 4 + k/6 = x times 4
loris [4]
Here is your answer....

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
Find csc(angle) if tan(angle) = -4/3 and sin(angle) is positive.
Lerok [7]

Answer:

hm?

Step-by-step explanation:

hm

7 0
3 years ago
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