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IRISSAK [1]
3 years ago
9

39 is what percent of 156?

Mathematics
2 answers:
marshall27 [118]3 years ago
6 0
The answer Is 25% of 156
sashaice [31]3 years ago
4 0
Since 39/156 can be simplified to 1/4, 1/4 of 100% is 25%.
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You are skateboarding at a pace of 30 meters every 5 seconds. your friend is in-line skating at a pace of 9 meters every 2 secon
AnnyKZ [126]

In order to graph the relationship, we will need to write the expression as the equation of a straight line as shown:

d = mt + b

d is the distance covered

t is the time taken

m is the speed

If you are skateboarding at a pace of 30 meters every 5 seconds. your friend is in-line skating at a pace of 9 meters every 2 seconds, this can be written as (5, 30) and (2, 9)

Get the slope of the line:

m = (9-30)/(2-5)

m = -21/-3

m = 7

Substitute m = 7 and the coordinate (2, 9) into the equation y = mt + b

9 = 7(2) + b

9 = 14 + b

b = -5

The required equation to plot will be expressed as y = 7t - 5

Plot the required graph

Learn more here: brainly.com/question/17003809

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

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

<u>So, 95% confidence interval for the true mean, </u>\mu<u> is ;</u>

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

<u>95% confidence interval for</u> \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

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

<u>So, 98% confidence interval for the true mean, </u>\mu<u> is ;</u>

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

<u>98% confidence interval for</u> \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
Help i forgot 3(x−4)≥−2x+3
alekssr [168]

Answer:

The answer is x_> 3

Step-by-step explanation:

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3 years ago
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The weather in Montana
Phantasy [73]

Answer:

13 degrees warmer

Step-by-step explanation:

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2 years ago
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A pond is being dug according to plans that have a scale of 3 inches = 9 feet. What is the scale factor of the Plans?
liq [111]

Answer:

1/36 scale factor

Step-by-step explanation:

3 inches/108 in

=1/36 scle factor

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3 years ago
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