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vodka [1.7K]
4 years ago
7

(a) A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of g

as (therms) used during the month of January is determined for each house. The resulting observations are 125, 103, 118, 109, 122, 82, 99, 138, 151, 149. Let μ denote the average gas usage during January by all houses in this area. Compute a point estimate of μ.
Mathematics
1 answer:
rewona [7]4 years ago
5 0

Answer:

In average, houses in the particular area use 119,6 therms of gas during the month of January.

Step-by-step explanation:

The μ formula is:

μ= ΣXi/N

ΣXi= is the sum of each xi. xi is each observation in the sample.

N= Total number of observations.

For this case:

ΣXi= 125+103+118+ 109+ 122+ 82+ 99+ 138+ 151+ 149

ΣXi= 1196

N= 10

μ= 1196/10

μ= 119,6

In average, houses in the particular area use 119,6 therms of gas during the month of January.

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Which of the following equations represents a line that is perpendicular to y=3x+11 and intercepts the line at the point, (-2, 5
yaroslaw [1]
If the line is perpendicular the slope is the reciprocal so it would be
Y=-1/3x
Then if the line has to go through the point (-2,5) the y-intercept must be 5 making the answer
Y=-1/3x+5
7 0
3 years ago
A jeweler is making a triangular pendant. He has two sides of length 23 mm and 31 mm. What are the
denis-greek [22]

Answer:

38.60mm

Step-by-step explanation:

Step one:

Given data

We are given that the dimension of the triangles are length 23 mm and 31 mm

Let us assume that the triangle is a right angle triangle

Step two:

Applying the Pythagoras theorem we can find the third as

z^2= x^2+y^2\\\\z^2= 31^2+23^2\\\\z^2= 961+529\\\\z^2= 1490\\\\

square both sides

z= √ 1490

z= 38.60mm

Hence a possible dimension of the third side is 38.60mm

3 0
3 years ago
The riding stables just received an unexpected rush of registrations for the next horse show, and quickly needs to create some a
liraira [26]

Answer:

Maximum added area will be 60,000 ft²

Each paddock measure 200 ft by 150 ft, and the paddocks share a 200-ft long side.

Step-by-step explanation:

Consider the provided information.

There is sufficient funding to rent 1200 feet of temporary chain-link fencing.

The plan is to form two paddocks with one shared fence running down the middle.

The total length of fencing is 1200 feet that is same as the total perimeter including that shared line down the middle.

Therefore, the perimeter is

P = 1200 = 2L + 3w

Solve for L

1200-3w = 2L

L=600-\frac{3w}{2}

As we know the area of rectangle is A=L×W

Substitute the value of L in above formula.

A=(600-\frac{3w}{2})\times w\\A=600w-\frac{3w^2}{2}

A=-\frac{3w^2}{2}+600w

The above equation makes a downward parabola.

In order to find the maximum, find the vertex of parabola as shown:

W=\frac{-b}{2a}=\frac{600}{2\times\frac{3}{2}}=200

Hence, the width should be 200 ft in order to get maximum area,

Therefore, the length will be:

1200 = 2L + 3(200)

L=300

Now substitute the value of L and W in area formula.

A=300×200=60,000

Hence, maximum added area will be 60,000 ft²

Hence, the length should be 300 ft, and each paddock should then be 150 feet long.

Each paddock measure 200 ft by 150 ft, and the paddocks share a 200-ft long side.

5 0
3 years ago
Help me to simplify; <br> log_{5} 12.5+log_{5}10
liq [111]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
A bacterial culture starts with 400 bacteria and grows at a rate proportional to its size. After 2 hours there will be 800 bacte
kotykmax [81]

Answer:

  1. P(t) = 400×2^(t/2)
  2. 6400
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Step-by-step explanation:

The wording "a rate proportional to its size" is an indication that growth (or decay) is exponential.

1) For many problems involving exponential growth, I like to use the numbers given in the problem statement as follows.

  population = (initial population) × (growth factor)^(t/(growth period))

where the "growth period" is the period of time in which the population is multiplied by the "growth factor".

Here, we're given ...

  (initial population) = 400

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  (growth period) = 2 . . . . hours

So, our population function can be written as ...

  P(t) = 400×2^(t/2)

__

2) Putting t=8 into the formula, we get ...

  P(8) = 400×2^(8/2) = 400×16 = 6400

After 8 hours, the population will be 6400.

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3) Fill in the given number and solve for t.

  2270 = 400×2^(t/2)

  2270/400 = 2^(t/2)

Taking logs, we have ...

  log(227/40) = (t/2)log(2)

  t = 2×log(227/40)/log(2) ≈ 5.009

After 5.0 hours, the population will reach 2270.

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