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Soloha48 [4]
3 years ago
9

If f(x)=2x-5, find a. f(4) b. f(-3)

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0

Replace x in the given equation with each value:

a. f(4) = 2(4) -5

= 8-5

=3

b. (f-3) = 2(-3) - 5

= -6 -5

= -11

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Let A and B are n x n matrices from which A is invertible. Suppose AB is singular. What conclusion can be made about the inverti
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Answer: Matrix B is non- invertible.

Step-by-step explanation:

A matrix is said to be be singular is its determinant is zero,

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Given :  A and B are n x n matrices from which A is invertible.

Then A must be non-singular matrix.                            ( from 2 )

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Suppose you want to eat lunch at a popular restaurant. The restaurant does not take reservations, so there is usually a waiting
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Answer:

a) z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{15-19}{4} =-1

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-1)=1-P(z

And replacing we got:

P(X >20| X>15)= \frac{0.401}{0.841}= 0.477

b) z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{18-19}{4} =-0.25

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-0.25)=1-P(z

And replacing we got:

P(X >20| X>18)= \frac{0.401}{0.599}= 0.669

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the length of time waiting to be seated of a population, and for this case we know the distribution for X is given by:

X \sim N(19,4.0)  

Where \mu=19 and \sigma=4.0

Part a

For this cae we want to find this probability:

P(X >20| X>15)

And if we use the conditional probability formula we got:

P(X >20| X>15)= \frac{P(X >20 \cap X>15)}{P(X>15)}=\frac{P(X>20)}{P(X>15)}

We can solve the problem using the z score formula given by:

z = \frac{x-\mu}{\sigma}

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{15-19}{4} =-1

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-1)=1-P(z

And replacing we got:

P(X >20| X>15)= \frac{0.401}{0.841}= 0.477

Part b

P(X >20| X>18)= \frac{P(X >20 \cap X>18)}{P(X>18)}=\frac{P(X>20)}{P(X>18)}

We can solve the problem using the z score formula given by:

z = \frac{x-\mu}{\sigma}

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{18-19}{4} =-0.25

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-0.25)=1-P(z

And replacing we got:

P(X >20| X>18)= \frac{0.401}{0.599}= 0.669

5 0
4 years ago
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