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AVprozaik [17]
3 years ago
11

Please help i will marl you brainless tha thangkyou​

Mathematics
1 answer:
pogonyaev3 years ago
8 0

Answer:

5

Step-by-step explanation:

The faces are 5

And the Name is rectangle prism

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How to find y-intercept of 2x - 5y + 15​
irinina [24]

Answer:

y-intercept = 3

Step-by-step explanation:

  • y = mx + c
  • rearrange 2y-5y+15
  • let y as a subject
  • c = y-intercept

7 0
4 years ago
If k is the midpoint of JL, JK =8x + 11 and KL =14x-1, find JL
katen-ka-za [31]
J------------K------------L
   8x+11        14x-1


<span>If K is the midpoint of JL then JK = KL
8x+11 = 14x-1
11 + 1 = 14x - 8x
12 = 6x
x = 12/6
x = 2

JK = 8x + 11 = 8*2 + 11 = 16+11 = 27
KL = JK = 27

JL = JK + KL = 27 + 27 = 54</span>
3 0
4 years ago
Read 2 more answers
Can anyone help plzzz?
Rashid [163]
1 inch = 5 yards that’s the awnser
7 0
3 years ago
Read 2 more answers
Find the difference,<br> (6x + 18x²) - 3²<br> A 21x²<br> B. 21x3<br> c. 3x + 15 x²<br> D. 6x + 15x²
SashulF [63]
I think the answer is d
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
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