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AlekseyPX
3 years ago
13

Cari owns a horse farm and a horse trailer than can transport up to 8000 pounds of livestock and tack. She travels with 5 horses

whose combined weight is 6240 pounds. let t represent the average weight of tack per horse. which of the following inequalities could be used to determine the weight of tack Cari can allow for each horse?
A. 6240+t≤8000
B. 8000≥6240t
C. 6240+5t≤8000
D. 8000+6240≤5t
E. 8000-6240≤5t
Mathematics
2 answers:
horsena [70]3 years ago
8 0

Answer: your answer should be D

Step-by-step explanation:

because the trailer is already 8000 pounds. so you would add that weight with the weight of the horses. the 5t would represent the number of horses and how many pounds they are since you don't know that part of the question.

Hdbdbfbfbfbf2 years ago
0 0

I think idk it’s b isjdnd

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a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample s
Alex

Answer:

95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

Step-by-step explanation:

We are given that a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97.

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

                         P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} }  ~ \chi^{2}__n_-_1

where, s^{2}  = sample variance = 97^{2} = 9409

              n = sample of boxes = 15

           \sigma^{2}  = population variance

<em>Here for constructing 95% confidence interval we have used chi-square test statistics.</em>

So, 95% confidence interval for the population variance, \sigma^{2} is ;

P(5.629 < \chi^{2}__1_4 < 26.12) = 0.95  {As the critical value of chi-square at 14

                                         degree of freedom are 5.629 & 26.12}  

P(5.629 < \frac{(n-1)s^{2} }{\sigma^{2} } < 26.12) = 0.95

P( \frac{5.629 }{(n-1)s^{2} } < \frac{1}{\sigma^{2} } < \frac{26.12 }{(n-1)s^{2} } ) = 0.95

P( \frac{(n-1)s^{2} }{26.12 } < \sigma^{2} < \frac{(n-1)s^{2} }{5.629 } ) = 0.95

<u><em>95% confidence interval for</em></u> \sigma^{2} = [ \frac{(n-1)s^{2} }{26.12 } , \frac{(n-1)s^{2} }{5.629 } ]

                                                  = [ \frac{14 \times 9409 }{26.12 } , \frac{14 \times 9409 }{5.629 } ]

                                                  = [5043.11 , 23401.31]

Therefore, 95% confidence interval for the population variance of the number of tissues per box is [5043.11 , 23401.31].

6 0
3 years ago
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jek_recluse [69]

Answer:

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Step-by-step explanation:

it is real, and it is rational. It isn't irrational, and it is not a natural number either

6 0
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Answer:

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Step-by-step explanation:

Consider the provided information.

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Here n is 125 and the probability of late is 24 or q = 0.24

Thus substitute the respective values in the above formula.

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Hence, the 30 flights are expected to be late.

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