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svetlana [45]
3 years ago
14

housing prices in a certain neighborhood average at 97.86 per square foot. If one house in this neighborhood is 1400 square feet

, what should it be priced at?
Mathematics
2 answers:
Arturiano [62]3 years ago
8 0
The average price of such a house is
  (97.86/ft²)×(1400 ft²) = 97.86×1400 = 137,004
Alina [70]3 years ago
5 0
The average price is 137004$ for 1400 square feet
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1/2(x + 5)2 + 2 = 42.5
Monica [59]

Answer:

x=35.5

Step-by-step explanation:

reduce numbers

remove parentheses

add the numbers

move constant to the right

subtract numbers

7 0
3 years ago
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Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
2 years ago
What is 24/28 in simplest form?!?
vovikov84 [41]
Divide top and bottom by four.
6/7 is the answer
3 0
2 years ago
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An art exhibit is made by stacking three identical prisms on top of each other end-to-end so their smallest faces overlap. One o
cestrela7 [59]

The surface area of the exhibit is 1,638 cubic inches.

Step-by-step explanation:

Step 1:

If three prisms are out on top of each other we get a shape as follows;

The length of the base = 9 inches,

The width of the base =6 inches, and

The height of the exhibit = 3(17) = 51 inches.

To find the surface area of the total exhibit we consider there to be only one shape which the above dimension.

Step 2:

There are 6 faces to the exhibit.

The surface area of a rectangular prism is given by;

A=2(w l+h l+h w)

By substituting the values, we get

A=2((6)(9)+(51) (9)+(51) (6)) = 2 (819) = 1,638.

The surface area of the exhibit is 1,638 cubic inches.

4 0
3 years ago
Solve using system of equations
IgorC [24]

Answer:

x=8

Step-by-step explanation:

set both equations equal to each other

x^2-3x-10=-2x=2

combine like terms

x^2=1x+8

x+8

Hope this helps!

8 0
2 years ago
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