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andre [41]
3 years ago
6

Q. Friday night’s hockey game was attended by 180 students. This is about 60 % of the school’s population. What is the total pop

ulation of the school?
Mathematics
1 answer:
OlgaM077 [116]3 years ago
7 0

Answer:

The total population of the school is 300 students.

Step-by-step explanation:

Considering, we are working with percentage we can use the equation:

y = %x.

To fill in the equation you can use what you already know, 60% = 180

180 = 0.60x

To solve this, you can then divide both sides of the equation by 0.60. This will then give you your answer, 300 = x.

For checking, you can do 180 = 0.60(300) which will result in 180 = 180.

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What is the solution to this inequality?<br> –16x &gt; –80
Kisachek [45]

Answer:

x < 5

Step-by-step explanation:

–16x > –80

Divide each side by -16, remembering to flip the inequality

-16x/-16< -80/-16

x < 5

7 0
3 years ago
Suppose a simple random sample of size nequals 150 is obtained from a population whose size is Upper N equals 30 comma 000 and w
denis23 [38]

(a) Correct answer is Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10.

(b) The value of P (X ≥ 770) is 0.0143.

(c) The value of P (X ≤ 720) is 0.0708.

Let X = number of elements with a particular characteristic.

The variable p is defined as the population proportion of elements with the particular characteristic.

The value of p is:

p = 0.74.

A sample of size, n = 1000 is selected from a population with this characteristic.

(a)

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

               μ = p

The standard deviation of this sampling distribution of sample proportion is:

                  σ = \sqrt \frac{p(1-p)}{n}

The sample selected is of size, n = 1000 > 30.

Thus, according to the central limit theorem the distribution of  is Normal, i.e. .

p~ N(μ = 0.74, σ =0.0139)

Thus the correct option is (A).

(b) We need to compute the value of P (X ≥ 770).

Apply continuity correction:

P (X ≥ 770) = P (X > 770 + 0.50)

                  = P (X > 770.50)

Then,

   p > 770.5/1000 = 0.7705

Compute the value of  P( p > 0.7705) as follows:

P( p > 0.7705) = P(p -μ/σ > 0.7705 - 0.74/0.0139)

                       = P( Z > 2.19)

                       = 1 - P( Z< 2.19)

                       = 1 - 0.98574

                       = 0.01426

                       ≈ 0.0143

Thus, the value of P (X ≥ 770) is 0.0143.

(c)

We need to compute the value of P (X ≤ 720).

Apply continuity correction:

P (X ≤ 720) = P (X < 720 - 0.50)

                  = P (X < 719.50)

Then

Compute the value of  as follows:

P( p < 0.7195) = P(p -μ/σ > 0.7705 - 0.74/0.0139)

                       = P(Z < - 1.47)

                       = 1 - P(Z < 1.47)

                       = 1 - 0.92922

                       = 0.07078

                        ≈ 0.0708

Thus, the value of P (X ≤ 720) is 0.0708.

Learn more about Simple Random sample:

brainly.com/question/13219833

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