Which expression is equivalent to 6p + 6p? A. 12p B. 36p C. 12 + 2p D. p6

A clock radio is on sale for 25% off it’s original retail price of $30. If the delivery charge for the radio is 10% if its purch

Alex17521 [72]

To buy the radio it would cost $22.5 I believe

To have it delivered it would cost $2.5 I believe

So overall that would cost $25

( I apologize if I'm wrong)

3 0

4 years ago

How do you find the area of this regular polygon

Maksim231197 [3]

Answer:

By using the formula of the octagon:

10m=1000cm

A ≈ 4.83×10^6 cm²

5 0

3 years ago

SOMEONE PLZ HELP ME The two box-and-whisker plots below show the scores on a math exam for two classes. What 2 things do the int

Savatey [412]

A box-and-whisker plot shows 5 number summary: Class A Class B1) minimum value 65 562) 1st quartile 66 793) median 81 914) 3rd quartile 89 945) maximum value 95 100
Interquartile range = 3rd Quartile - 1st Quartile Class A: 89 - 66 = 23Class B: 94 - 79 = 15
IQR indicates the spread of the of the scores over the median. Class A has a spread that his more than 50% of the scores of Class B.
Hope this helps!! :)

3 0

3 years ago

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

grin007 [14]

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:

$X \sim N(19,4.0)$