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Schach [20]
4 years ago
8

Jaden and his family are using a map to drive down to Napa Valley. On the map, the distance is given by the equation c = 1.3m, w

here c is the number of centimeters on the map and m is the corresponding real-world distance in miles. If they cover a distance of 18 miles, that would mean a distance of how many centimeters on the map.

Mathematics
2 answers:
Burka [1]4 years ago
8 0

Answer:

23.4

Step-by-step explanation:

Kisachek [45]4 years ago
3 0

Well if C=1.3m , we substitue 18 into M and get C=1.3(18)= 234

so 234cm

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Elina [12.6K]
This is very simple to do. Take out your calculator and insert 14.4. However, move the decimal two places to the left to get .144, and multiply it by 72.5 to get your answer of 10.44.
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3 years ago
The amounts of a sum of money in 2 years and 5 years are rs 2800 and e 3250respectvely .find the sum
sammy [17]

Answer:

money : x

in 2 years : x+2

in 5years: x+2= 2800 + 3250

sum= x+2= 6050

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x=3025

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3 years ago
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umka21 [38]

Answer:

answer is c.

Step-by-step explanation:

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8 0
3 years ago
The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand
Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

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 mean strength of 50 items = 74.28

            \sigma = population standard deviation = 2.15

            n = sample of items = 50

            \mu = population mean tensile strength after machine was adjusted

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

So, 95% confidence interval for the population 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

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } } ]

                 = [ 74.28-1.96 \times {\frac{2.15}{\sqrt{50} } } , 74.28+1.96 \times {\frac{2.15}{\sqrt{50} } } ]

                 = [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

<em>Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.</em>

8 0
4 years ago
WILL GIVE BRAINLIEST IF CORRECT
Mazyrski [523]

Answer:

E

Step-by-step explanation:

6 0
3 years ago
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