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kherson [118]
4 years ago
12

Which pair represents equivalent ratios? A Begin fraction . . . 2 over 3 . . . end fraction . . . and . . . begin fraction . . .

9 over 15 . . . end fraction B Begin fraction . . . 5 over 8 . . . end fraction . . . and . . . begin fraction . . . 15 over 21 . . . end fraction C Begin fraction . . . 3 over 12 . . . end fraction . . . and . . . begin fraction . . . 6 over 18 . . . end fraction D Begin fraction . . . 4 over 10 . . . end fraction . . . and . . . begin fraction . . . 12 over 30 . . . end fraction
Mathematics
2 answers:
Blizzard [7]4 years ago
5 0

Equivalent ratios are those ratios which are same that is on simplifying each of them, we will get the same answer.

A.\frac{2}{3} and \frac{9}{15} = \frac{3}{5}

So they are not equivalent

B.\frac{5}{8} and \frac{15}{21} = \frac{5}{7}

So they are not equivalent

C.\frac{3}{12} and \frac{6}{18} = \frac{1}{4} and \frac{1}{3}

They are not equivalent too.

D.\frac{4}{10} and \frac{12}{30} = \frac{2}{5} and \frac{2}{5}

And they are equivalent. SO the correct option is D .

artcher [175]4 years ago
3 0

Answer:

the answer is D

Step-by-step explanation:

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Part A: Find the LCM of 7 and 12. Show your work. (3 points)
inna [77]

Answer:

A. 84; B. 8; C. 8 × 19

Step-by-step explanation:

Part A. Least common multiple

Step 1. List the prime factors of each.

7 = 7

12 = 2 × 2 × 3

Step 2: Multiply each factor the greatest number of times it occurs in either number.

7 has one 7; 12 has two 2s and one 3.

LCM = 7 × 2 × 2 × 3

LCM = 7 × 12

LCM = 84

Part B. Highest common factor

Find all the factors of 56 and 96.

Factors of 56: 1, 2, 4,     7, 8,      14,          28

Factors of 96: 1, 2, 4, 6,     8, 12,    16, 24,     32, 48

The highest factor that in both 56 and 96 is 8.

Part C. Factoring

56 + 96 = 8(7 + 12) = 8 × 19

The GCF is 8.

19 = 7 + 12 is the sum of two numbers that do not have a common factor.

4 0
4 years ago
When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
HACTEHA [7]

Answer:

(a) The value of P (X ≤ 2) is 0.8729.

(b) The value of P (X ≥ 5) is 0.0072.

(c) The value of P (1 ≤ X ≤ 4) is 0.7154.

(d) The probability that none of the 25 boards is defective is 0.2774.

(e) The expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defective boards.

The probability that a circuit board is defective is, <em>p</em> = 0.05.

The sample of boards selected is of size, <em>n</em> = 25.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={25\choose x}0.05^{x}(1-0.05)^{25-x};\ x=0,1,2,3...

(a)

Compute the value of P (X ≤ 2) as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

P(X\leq =x)=\sum\limits^{2}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.2774+0.3650+0.2305\\=0.8729

Thus, the value of P (X ≤ 2) is 0.8729.

(b)

Compute the value of P (X ≥ 5) as follows:

P (X ≥ 5) = 1 - P (X < 5)

              =1-\sum\limits^{4}_{x=0}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=1-0.9928\\=0.0072

Thus, the value of P (X ≥ 5) is 0.0072.

(c)

Compute the value of P (1 ≤ X ≤ 4) as follows:

P (1 ≤ X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

                   =\sum\limits^{4}_{x=1}{{25\choose x}0.05^{x}(1-0.05)^{25-x}}\\=0.3650+0.2305+0.0930+0.0269\\=0.7154

Thus, the value of P (1 ≤ X ≤ 4) is 0.7154.

(d)

Compute the value of P (X = 0) as follows:

P(X=0)={25\choose 0}0.05^{0}(1-0.05)^{25-0}=1\times 1\times 0.277389=0.2774

Thus, the probability that none of the 25 boards is defective is 0.2774.

(e)

Compute the expected value of <em>X</em> as follows:

E(X)=np=25\times 0.05=1.25

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{25\times 0.05\times (1-0.05)}=1.09

Thus, the expected value and standard deviation of <em>X</em> are 1.25 and 1.09 respectively.

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