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4 years ago

6

Question: (5x-8)/(45x^2--42x-48)

2 answers:

Vladimir79 [104]4 years ago

7 0

Answer:

D

Step-by-step explanation:

9x+6

Edge 2020

My name is Ann [436]4 years ago

4 0

To solve this problem you must apply the proccedure shown below:

1. You have the folllowing expression given in the problem above:

$(5x-8)/(45x^{2}-42x-48)$

2. The first thing you must do is to factor the denominator, as following:

$(5x-8)/3(3x+2)(5x-8)$

3. Now, simplify it and apply the distributive property in the denominator:

$1/3(3x+2)=1/9x+6$

The answer is: $1/9x+6$

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4(3)(-2)
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Jacob makes a scale drawing for a crate with a scale

olga55 [171]

Answer:

6 inches long and 4 inches wide

Step-by-step explanation:

The dimensions of the actual crate are 20 times those on Jacob's drawing. The dimensions on the builder's drawing are 1/10 of those, so (1/10)(20) = 2 times the dimensions on Jacob's drawing.

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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

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

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

=> $n = 5^2$

=> $n = 25$

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The answer is 0 because it isn't above 49 and therefore can't be rounded to 100.

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