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diamong [38]
3 years ago
11

Subtracting unlike fractions 8/9 - 5/6

Mathematics
1 answer:
inn [45]3 years ago
7 0
Hello,

Here is your answer:

The proper answer to your question is "1/18".

Here is how:

8*6=48
9*6=54
-
5*9=45
6*9=54

48-45=3

Which is 3/54!

3/3=1
54/3=18

Which means your answer is 1/18!

If you need anymore help feel free to ask me!

Hope this helps!
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Answer:

p > 5/14

Step-by-step explanation:

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Select the correct answer.
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The correct answer is A
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The area of a rectangle is 1,036 cm and its length is 74 cm.
Nikitich [7]

Answer:

(i) Breadth = 14 cm

(ii) Perimeter = 176 cm

Step-by-step explanation:

Given,

  • Area of a rectangle = 1036 cm²
  • Length = 74 cm

(i) Breadth = ?

We know that, Area of a rectangle = length × breadth.

Then,

1036 = 74 \times Breadth\\1036  \div 74 = Breadth\\\boxed{\bf\:14 \:cm = Breadth }

(ii) Perimeter of a rectangle = 2 (length + breadth)

Then,

Perimeter = 2(74 + 14)\\Perimeter = 2(88)\\\boxed{\bf\: Perimeter = 176 \: cm}

\rule{150pt}{2pt}

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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

#SPJ4

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