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4 years ago

Ms.Lopez has created a floor plan of a dollhouse. The area of the entire dollhouse is 576 square inches

Mathematics

1 answer:

harina [27]4 years ago

3 0

Answer:

(a)Area of the bedroom=96 square inches

(b)Area of the living room =144 square inches

(c)Area of the dollhouse that is not of the bedroom or living room =336 square inches.

Step-by-step explanation:

The area of the bedroom is 1/6 the area of the entire dollhouse

The area of the living room is 3/2 times the area of the bedroom.

The area of the entire dollhouse is 576 square inches

Let the area of bedroom=b

Let the area of living room=l

(a) The area of the bedroom is 1/6 the area of the entire dollhouse

$b=\frac{1}{6}X576=96$ square inches

(b)The area, in square inches, of the living room

The area of the living room is 3/2 times the area of the bedroom

l= $\frac{3}{2}X96=144$ square inches

(c)The total area of the house =576 square inches

Area of the bedroom=96 square inches

Area of the living room =144 square inches

Area of the dollhouse that is not of the bedroom or living room

=576-(96+144)=576-240=336 square inches.

The area of the living room is 3/2 times the area of the bedroom

This is gotten by subtracting the area of the bedroom and living room from the total area of the house.

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Answer:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

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 heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(63.6,2.5)$