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BigorU [14]
2 years ago
5

In a race , Ram covers 5 km in 20 min. How much distance will he cover in 100 min

Mathematics
2 answers:
Mekhanik [1.2K]2 years ago
5 0
You can use proportion
Km Min
5 20
? 100

(5*100)/20
500/20
=25
masya89 [10]2 years ago
4 0

Answer:

25 km

Step-by-step explanation:

100/20=5

5*5=25

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The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard de
inna [77]

Answer:

<em>The probability of spending between 4 and 7 days in recovery</em>

<em>P(4≤x≤7) = 0.5445</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given mean of the Population μ = 5.3 days

Given standard deviation of the population 'σ' = 2 days

Let 'X' be the random variable in normal distribution

Let    x₁ = 4

Z_{1} = \frac{x_{1}-mean }{S.D} = \frac{4-5.3}{2} = -0.65

Let    x₂ = 7

Z_{2} = \frac{x_{2}-mean }{S.D} = \frac{7-5.3}{2} = 0.85

<u><em>Step(ii):-</em></u>

<u><em>The probability of spending between 4 and 7 days in recovery</em></u>

<em>P(4≤x≤7) = P(-0.65≤Z≤0.85)</em>

<em>              =  P(Z≤0.85) - P(Z≤-0.65)</em>

<em>             = 0.5 + A( 0.85) - ( 0.5 - A(-0.65) </em>

<em>             = 0.5 + A( 0.85) -  0.5 +A(0.65)   ( ∵A(-0.65) = A(0.65)</em>

<em>            =   A(0.85) + A(0.65)</em>

<em>           = 0.3023 + 0.2422</em>

<em>          = 0.5445</em>

<u><em>Final answer:</em></u><em>-</em>

<em>The probability of spending between 4 and 7 days in recovery</em>

<em>P(4≤x≤7) = 0.5445</em>

<em>            </em>

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Step-by-step explanation:

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