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3 years ago

Geumath malah 17.pdf

Mathematics

2 answers:

Ymorist [56]3 years ago

8 0

Answer:

He is wrong.

In my answer I found the complete volume of the pool and then what the volume would be 10 cm empty. You probably only need to find the 10 cm emptied to answer, just a heads up. if you have any questions though let me know.

Step-by-step explanation:

To most easily figure out the volume of the empty pool I think it is easiest to break it up into four parts..

First break off that skinny end bit so it will have 1 m in height, 4 m in length and 5m in width. Width is never going to shange because of how we are breaking this up, and the width is the distance from you into the paper, if that makes sense. Basically not up and down or left and right.

Next another rectangular prism from the right side with 2 height, 2m length and again 5m width.

Now you have a trapezoidal prism with one base of length 1m, another of base 2m and a height of length 4m.

To fond the volume of prisms you want to find the area of their bases then multiply it by their heights. In this instance height is going to be the width again, and the bases are th parts facing us on the picture.

For a rectangular prism area of the base is easy, just multiply length by height. a trapezoid is taking the two bases (b1 and b2) ading them together and dividing that sum by 2, THEN taking this new number and multiplying it by what would be the height of the trapezoid. the trapezoid has its bases as two heights, and its height then would be the horizontal length of the pool. Sorry if that is confusing. Here are the three volumes though.

Rectangular 1: 4*1*5 = 20 m^3

Rectangular 2" 2*2*5 = 20m^3

Trapezoidal: = ([1+2]/2)*4*5 = 30 m^3

So total it is 70 m^3

The question then says each cubic meter can contain 1,000 liters of water. 70 m^3 then is 70,000 liters

The question also says there are three barrels of 20,000 liters each, so combined that's 60,000 liters.

Finally it wants to know if Sam is right saying if you dump all of the 60,000 liters into the pool the surface of the water will not reach the top of the pool by 10 cm. 60,000 is less than 70,000, but we don't know how much lower it is in cm.

The trick here is to know that we are lowering a specific dimension of each prism to a certain amount to be lower by 10cm

In the 4x1x5 rectangular prism the 1m side is lowering.

In the 2x2x5 rectangular prism the 2m side representing the vertical length is getting some amount taken away.

int he trapezoidal prism both of the "bases" are getting an amount taken.

So the trick here is to set up the math again, except this time with 10 cm less being subtracted from each part, then solving t and see if it gets us the 60,000 liters or 60 m^3. Since we are measurng in meters, 10 cm less is the same a subtracting .1, or if you prefer subtracting a decimeter.

4*1*5 becomes 4(.9)5 = 18

2*2*5 becomes 2(1.9)5 = 19

Trapezoidal: = ([1+2]/2)*4*5 becomes ([(.9)+(1.9)]/2)*4*5 = 28

Adding this all together gets us 65 or 65,000 liters. This means that if the pool is filled 10 cm from the top the volume would be 65,000 liters, still more than if the three tankers are emptied completely into it.

You could have probably just done the second part, where we sbtracted 10 cm from each of the dimensions, but I am going to leave it all since I wrote it.

Alex17521 [72]3 years ago

8 0

9514 1404 393

Answer:

Sam is not correct.

Step-by-step explanation:

As shown in the attachment, the "base" (front face) of this prism can be divided into a rectangle and a trapezoid. The relevant area formulas are ...

A = LW . . . . area of a rectangle of length L and width W

A = 1/2(b1 +b2)h . . . . area with parallel bases b1 and b2 and height h

The rectangle portion of the "base" of the pool has area ...

A = (10 m)(1 m) = 10 m²

The trapezoid portion of the "base" of the pool has area ...

A = 1/2(6 m +2 m)(1 m) = 4 m²

Then the "base" area is ...

B = 10 m² +4 m² = 14 m²

The volume of the prism is given by ...

V = Bh . . . . where B is the base area and h is the distance between bases

V = (14 m²)(5 m) = 70 m³

We are told that 1 m³ = 1000 L and that 3 tankers of 20,000 L each are emptied into the pool. That leaves an unfilled volume of ...

70 m³ -3×(20 m³) = 10 m³ . . . . unfilled volume

The surface area of the pool is ...

A = LW = (10 m)(5 m) = 50 m²

So the unfilled volume has a height of ...

V = Bh

10 m³ = (50 m²)h

h = (10/50) m = 20/100 m = 20 cm . . . . . the height of the unfilled space

The water in the pool will be 20 cm below the top, not 10 cm. Sam is not correct.

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3 years ago

Today there are a total of six toll-free area codes: 800, 844, 855, 866, 877, and 888. Assume that all seven digits for the rest

Ludmilka [50]

Answer:

$6 \times 10^{7}$

Step-by-step explanation:

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Finding the possible toll free numbers for one area code and multiplying that by 6 will give use the total number of toll free numbers for all 6 area codes.

Considering: 800-abc-defg

The first number "a" can take any digit from 0 to 9. So there are 10 possibilities for this place. Similarly, the second number can take any digit from 0 to 9, so there are 10 possibilities for this place as well and same goes for all the 7 numbers.

Since, there are 10 possibilities for each of the 7 places, according to the fundamental principle of counting, the total possible toll free numbers for one area code would be:

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4 0

3 years ago

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago

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agasfer [191]

5032.34
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5 0

3 years ago

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solmaris [256]

Answer:

Step-by-step explanation:

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3 years ago