The relationship between the number of hours a plumber works and the total work fee

Mathematics

2 answers:

crimeas [40]2 years ago

7 0

Answer:

The answer is B,C, And D

Step-by-step explanation:

I just saw it on Khan and got it right.

Lynna [10]2 years ago

5 0

Answer:

ANSWERS ARE B, C, AND D.

Step-by-step explanation:

I found the unit rate of 350 and 5, because we can use a calculator (as stated on top) I just punched in 350 divided by 5 and got 70. Now i know this may not sound that "professional", but nothing needs to always be professional as long as its efficient and most importantly gets you the right answer. I then took the time to multiply each number, (the ones on the left : 3, 3.5, 6, 7.25, and 8) by 70. If the answer you got on the calculator match up to the number on the right, (240, 420, 507.50, and 490) its correct, if it doesn't match up, you can tell that its most likely not the answer. You should end up with three answers that all end up having the unit rate of 70. Hope this helps anyone who needed it.

ANSWERS ARE B, C, AND D.

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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)