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soldi70 [24.7K]
3 years ago
6

What’s 18x3+12-1x34+13-2x56?

Mathematics
2 answers:
Alina [70]3 years ago
8 0

18 x 3 + 12 - 1 x 34 + 13 - 2 x 56 = 67

<=Work=>

18 x 3 = 54

1 x 34 = 34

2 x 56 = 112

(54) + 12 - (34) + 13 - (112)

66 - 34 + 13 - 112

32 + 13 - 112

45 - 112

= 67

Oliga [24]3 years ago
4 0

Answer: -67

Step-by-step explanation: Use PEMDAS, or order of operations. EPIC. Parenthesis, exponents, multiplication, division, addition, subtraction. Lit. Now, the first thing to do is multiply. 18 x 3=54.

54 + 12 - 1 x 34 + 13 - 2 x 56

Multiply 1 x 34 for 34

54 + 12 - 34 + 13 - 2 x 56

Multiply 2 x 56 to get 112

54 + 12 - 34 + 13 - 112

Now addition, so add 54 to 12 for 66

66 - 34 + 13 - 112

You also do the operation from left to right if there is addition and subtraction

66-34 = 32

32 + 13 -112

Add 13 to 32 for 45

45 - 112 = -67

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Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

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P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

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P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

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P(0 < z < 0.79) = 0.7852- 0.5

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e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

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From the standard normal distribution tables,

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P(-1.58 < z < 0 ) = 0.5 -  0.0571

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P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

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