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2 years ago

A rectangular pool is 30 1/3 feet long and 12 1/2 feet wide. What is the area of the pool?

Mathematics

1 answer:

Tomtit [17]2 years ago

4 0

You just need to multiply them together the. You'll get 2275/6 or 379 1/6 sq feet

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Can you please help this is urgent??!!!!

Thepotemich [5.8K]

Answer:

B. 4

Step-by-step explanation:

A direct proportion has the form

y = kx or k=y/x,

where k is the proportionality constant.

Let's test some points.

Second point

k = 8/2 = 4

Fifth point

k = 23/5.75 = 4

You will get the same result for every point, so the constant of proportionality is 4.

3 0

2 years ago

Find the sum of the first 100 terms of the sequence below: -19, -15, -11, -7, -3, ... S100

ch4aika [34]

Answer:

17900

Step-by-step explanation:

First term a = -19

common difference d = -15-(-19) = 4

number of terms, n = 100

sum of first 100 terms,

n/2(2a+(n-1)d)

= 100/2(2×(-19)+(100-1)4)

= 50×(-38+99×4)

= 50×358

= 17900

5 0

2 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)