San francisco is one of the most expensive cities in which to live in the united states. as of february 2013, the mean rent for a one-bedroom apartment in the mission district was $2600.6 assume that the distribution of rents is approximately normal and the standard deviation is $200. a one-bedroom apartment in the mission district is selected at random. (a) find the probability (±0.0001) that the rent is less than $2470. p(x< 2470) =

(b) find the probability (±0.0001)that the rent is between $2560 and $2640. p(2560⩽x⩽2640) =

(c) find a rent r (±0.01) such that 90% of all rents are less than r dollars per month. r =

a) 0.2569 or 25.69%

b) 0.1585 or 15.85%

c) $2,856.91

Step-by-step explanation:

a)  

Here we want to calculate the area between 0 and 2,470 (since no rent is negative) under the Normal curve of mean 2,600.6 and standard deviation 200.

We can do this easily with a spreadsheet and we get that the probability is 0.2569 or 25.69% (See picture 1)

b)

Now we look for the area under the curve between 2,560 and 2,640 which is 0.1585 or 15.85% (See picture 2)

c)

Here we are looking for a value r such that the area under the curve to the left of r is 90%=0.9.

Again, using the computer we find that value is r= $2,856.91

(See picture 3)

it is probably the graph on the bottom right.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15