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rusak2 [61]
2 years ago
7

ali's tyre pressure gauge shows a reading witch is 8 % lower than the actual pressure when the gauge shows 33.58

Mathematics
1 answer:
Rama09 [41]2 years ago
5 0

Answer:

the actual pressure is 31.09

Complete question:

What is the actual pressure when Ali's gauge shows 33.58

Step-by-step explanation:

Given,

Ali's tyre presser gauge show a reading 8% higher than the actual pressure.

Ali's gauge shows 33.58.

Let the actual pressure be x.

According to problem,

 x\times\frac{108}{100} =33.58 { [If 8% increasing  of a no ⇒ The number becomes =                      (\frac{100+8}{100})of the number}

x= \frac{33.58\times100}{108}

x=31.09

Therefore, the actual pressure is 31.09

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Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.
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Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

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Given that;

fx(x) = { 1/5 ; 50 < x < 100

              0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

                          1 ;   x > 100

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n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

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b)

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Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

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= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0
2 years ago
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