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Lemur [1.5K]
2 years ago
10

if five times a number added to 4 is divided by the number plus 6 the result is nine halves, Find The Number

Mathematics
1 answer:
Leto [7]2 years ago
7 0

Answer:

8/13

Step-by-step explanation:

Let the number be x

if five times a number added to 4 is divided by the number plus 6 the result is nine halves

  • five times the number = 5x
  • five times a number added to 4 = 5x+4
  • five times a number added to 4 is divided by the number = (5x+4)/x
  • five times a number added to 4 is divided by the number plus 6 = (5x+4)/x  + 6
  • nive halves = 9*(1/2)

five times a number added to 4 is divided by the number plus 6 the result is nine halves

(5x+4)/x  + 6 = 9(1/2)

=>(5x+4)/x  + 6 -6 = 9(1/2) -6 = 9/2-6 = (9-6*2)/2 = -3/2

=>(5x+4)= -3/2 *x= -3x/2

=>2(5x+4) = -3x

=> 10x +8 = -3x

=> 10x+3x = 8

=> 13x = 8

=> x = 8/13

Thus, the number is 8/13

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Step-by-step explanation:

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Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
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Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

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

From the standard normal distribution tables,

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

d.  z=0 and z=0.79

From the standard normal distribution tables,

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

Thus;

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

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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

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From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

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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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