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Sedaia [141]
3 years ago
14

The number of calories burned at the gym is normally distributed with a mean of 425 and a standard deviation of 51. Find the Z-s

core for each data value.
a.)268 b.)512 c.)450
Mathematics
1 answer:
Galina-37 [17]3 years ago
6 0

Answer:

(a) -3.708

(b) 1.706

(c) 0.490

Step-by-step explanation:

The z-score of normal distribution is given as:

z=\frac{x-\mu}{\sigma}

(a)

Given:

Score is, x=268

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{268-425}{51}=\frac{-157}{51}=-3.078

(b)

Given:

Score is, x=512

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{512-425}{51}=\frac{87}{51}=1.706

(c)

Given:

Score is, x=450

Mean value is, \mu =425

Standard deviation is, \sigma = 51

z-score is, z=\frac{450-425}{51}=\frac{25}{51}=0.490

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Answer:  7\frac{5}{7} unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

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∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

\triangle PDC \sim \triangle PAB

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\frac{PD}{PA} = \frac{DC}{AB}

\frac{PD}{PD+3} = \frac{11}{18} ( because PA = PD+DA)

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Again by the definition of similarity,

\frac{PC}{PB} = \frac{DC}{AB}

\frac{PC}{PC+7} = \frac{11}{18} ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA =  PD+DA = 33/7 + 3 = 7\frac{5}{7}

And, PB = PC + CB = 11 + 7 = 18


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