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____ [38]
3 years ago
9

Trucks are delivering gravel to a construction site each truck hold 7.5 cubic yards of gravel. The weight of 1 cubic yard of gra

vel is 1.48 tons. The gravel will be placed in containers that each hold 3.7 tons of gravel. how many contaoners of this size are needed to hold all the gravel from one truck
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0
The first thing you must do is to calculate the weight of the gravel (w) that the truck holds (7.5 yards³). So, you have:

 If the weight of 1 yard³ is 1.48 tons, then the weight of 7.5 yards³ is:

 w=7.5x1.48 tons
 w=11.1 tons

 The problem says that the gravel will be placed in containers that each hold 3.7 tons of gravel,so, If they need 1 cotainer to place 3.7 tons, how many containers they need to place 11.1 tons?

 1 container-------3.7 tons of gravel
                 x-------11.1 tons of gravel

 x=(11.1x1)/3.7
 x=11.1/3.7
 x=3 containers

 H<span>ow many containers of this size are needed to hold all the gravel from one truck?
</span>
 The answer is: 3 containers. 
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Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):
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Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, \mu_X represent the population mean for Brand B and let \mu_Y represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

        P.Q. = \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where, Xbar = Sample mean for Brand B data = 36.9

            Ybar = Sample mean for Brand A data = 32.9

              n_1  = Sample size for Brand B data = 13

              n_2 = Sample size for Brand A data = 9

              s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2}  }{n_1+n_2-2} } = \sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} } = 3.013

Here, s^{2}_X and s^{2} _Y are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < t_2_0 < 2.528) = 0.98

P(-2.528 < \frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}  } } < 2.528) = 0.98

P(-2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (Xbar -Ybar) -(\mu_X-\mu_Y) < 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

P( (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} < (\mu_X-\mu_Y) < (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} , (Xbar - Ybar) + 2.528 * s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} ]

[ (36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} , (36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9} ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

                     

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Step-by-step explanation:

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1 Kilobyte = 1,024 Bytes

1 Megabyte = 1,024 Kilobytes

1 Gigabyte = 1,024 Megabytes

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