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murzikaleks [220]
3 years ago
6

The mean lifetime of a tire is 4242 months with a variance of 4949. If 145145 tires are sampled, what is the probability that th

e mean of the sample would be greater than 42.842.8 months
Mathematics
1 answer:
SVEN [57.7K]3 years ago
6 0

The above question is not correctly written

Complete Question

The mean lifetime of a tire is 42 months with a variance of 49. If 145 tires are sampled, what is the probability that the mean of the sample would be greater than 42.8 months.

Answer:

0.083793

Step-by-step explanation:

We would be using the z score formula.

z score formula = z = (x - μ)/σ/√n

where

x is the raw score = 42.8

μ is the population mean = 42

σ is the population standard deviation =

In the above question, we were given variance = 49

Standard deviation = √Variance

= √49

= 7

n = number of samples = 145

z score = z = (x - μ)/σ/√n

= (42.8 - 42)/ (7/√145)

= 0.8/ 0.581318359

= 1.37618

Approximately to 2 decimal places= 1.38

Using the z score table to find the probability.

P(x ≤ 42.8) = P(z = 1.38) = 0.91621

P(x>42.8) = 1 - P(x<42.8)

1 - 0.91621

= 0.083793

Therefore, the probability that the mean of the sample would be greater than 42.8 is 0.083793

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In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t an
Alenkasestr [34]

Answer:  693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

A(t) =A_0e^{kt}

k = rate constant

t = time taken for decomposition = 1

A_0 = Initial amount of the reactant

A_t = amount of the reactant left =A_0-\frac{0.1}{100}\times A_0=0.999A_0

0.999A_0=A_0e^{k\times 1}

0.999=e^k

k=-0.001year^{-1}

for half life : t=t_\frac{1}{2}

A_t=\frac{1}{2}A_o

Putting in the values , we get

\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}

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8 0
4 years ago
Can someone help me please ​
andrey2020 [161]

Answer: y = 6

Step-by-step explanation:

A square's area can be done by using s^2, where s is y in this case.  Because there are 5 squares, the area of the figure is 5y^2.  Because the area is also 180cm, 5y^2=180.

Then divide both sides of the equation by 5 to get y^2 = 36.  Then square root both sides of the equation to get y = 6.

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3 years ago
What values of b satisfy 3(2b + 3)2 = 36?
KiRa [710]

the correct question is

What values of b satisfy 3(2b+3)^2 = 36


we have

3(2b+3)^2 = 36

Divide both sides by 3

(2b+3)^2 = 12

take the square root of both sides

( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3


b1=\frac{\sqrt{12}}{2} -\frac{3}{2}

b1=\sqrt{3} -\frac{3}{2}


b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}

b2=-\sqrt{3} -\frac{3}{2}

therefore


the answer is

the values of b are

b1=\sqrt{3} -\frac{3}{2}

b2=-\sqrt{3} -\frac{3}{2}


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3 years ago
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vredina [299]

Answer:

3

Step-by-step explanation:

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Answer:

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