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katovenus [111]

4 years ago

13

On a temperature versus time graph, how does the temperature at the beginning of a change of state compare with the temperature

at the end of the change? always lower always the same usually lower usually higher

1 answer:

vlada-n [284]4 years ago

5 0

Answer:

Always the same

Step-by-step explanation:

The temperature at the beginning of a change of state is always the same as the temperature at the end.

This is because phase change is an isothermal process. it means that all the energy absorbed during the phase change process is utilized in the breaking of the bonds in the compound as it changes from one state of matter to another.

As a result, no increase in the temperature of the material will be detected by the thermometer.

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3 years ago

The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and

Firdavs [7]

Answer:

1a

$P(39 < X < 48 ) = 0.8767$

1b

95% of all sample means will fall between $40.1 < \mu < 49.9$

1c

$\= x = 41. 795$

2

$n = 25$

Step-by-step explanation:

From the question we are told that

The mean is $n = 45$

The population standard deviation is $\sigma = 10$

The sample size is n = 16

Generally the standard error of the mean is mathematically represented as

$\sigma_{x} = \frac{ \sigma}{\sqrt{n} }$

=> $\sigma_{x} = \frac{ 10 }{\sqrt{16 } }$

=> $\sigma_{x} = 2.5$

Generally the probability that the sample mean will be between 39 and 48 minutes is

$P(39 < X < 48 ) = P( \frac{ 39 - 45}{ 2.5} < \frac{X - \mu }{\sigma } < \frac{ 48 - 45}{ 2.5} )$

=> $P(39 < X < 48 ) = P(-2.4 < Z< 1.2 )$

=> $P(39 < X < 48 ) = P( Z< 1.2 ) - P(Z < -2.4)$

From the z table the area under the normal curve to the left corresponding to 1.2 and -2.4 is

=> $P( Z< 1.2 ) = 0.88493$

and

$P( Z< - 2.4 ) = 0.0081975$

So

$P(39 < X < 48 ) = 0.88493 -0.0081975$

=> $P(39 < X < 48 ) = 0.8767$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * 2.5$

=> $E =4.9$

Generally the 95% of all sample means will fall between

$\mu -E < and \mu +E$

=> $45 -4.9\ and \ 45 + 4.9$

Generally the value which 90% of sample means is greater than is mathematically represented

$P( \= X > \= x ) = 0.90$

=> $P( \= X > \= x ) = P( \frac{\= X - \mu }{ \sigma_x} > \frac{\= x -45 }{ 2.5} ) = 0.90$

=> $P( \= X > \= x ) = P( Z > z ) = 0.90$

Generally from the z-table the critical value of 0.90 is

$z = -1.282$

$\frac{\= x -45 }{ 2.5} = -1.282$

=> $\= x = 41. 795$

Considering question 2

Generally we are told that the standard deviation of the mean to be one fifth of the population standard deviation, this is mathematically represented as

$s = \frac{1}{5} \sigma$

Generally the standard deviation of the sample mean is mathematically represented as

$s = \frac{\sigma }{ \sqrt{n} }$

=> $\frac{1}{5} \sigma = \frac{\sigma }{ \sqrt{n} }$

=> $n = 5^2$

=> $n = 25$

5 0

3 years ago