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3 years ago

12

What is the lenght in miles of the field

2 answers:

maw [93]3 years ago

8 0

(Perimeter/6)2= Length

nirvana33 [79]3 years ago

7 0

Ok this is what you do it went 12 miles and hour to travel the perimeter in 7.5 minuets there are 60 min in an hour do 12/60 thats 0.2 miles every min do 7.5x0.2=1.5 so the perimeter is 1.5 miles but you are looking for length so the length would be 0.5 because that mean the width would have to be 0.25 and

The length is 0.5miles

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Shtirlitz [24]

Answer:

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divide this:

18 ÷ 3 = 6 miles per gallon

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3 years ago

Read 2 more answers

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3 years ago

2. Barb buys 300 feet of wire, a crystal radio for $16.50, a flashlight for $7.95, and 2 batteries for $4.95 each. Her total bil

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3 years ago

In a certain lake, trout average 12 in. in length with standard deviation 2.75 in. and the bass average 4 lb. in weight with sta

NISA [10]

Answer:

The bass fish was the better catch

Step-by-step explanation:

From the question we are told that

The population mean for trout is $\mu_1 = 12 \ in$

The standard deviation is $\sigma_1 = 2.75 \ in$

The population mean for base is $\mu _2 = 4 \ lb$

The standard deviation is $\sigma_2 = 0.8 \ lb$

The number of trout caught $x_1 = 18$

The number of bass caught $x_2 = 6$

Generally z-value(standardized value ) for the of number trout caught is mathematically represented as

$z_1 = \frac{x_1 - \mu_1}{\sigma_1 }$

substituting value

$z_1 = \frac{18 - 12}{2.75 }$

$z_1 = 2.18$

Generally z-value(standardized value ) for the of number bass caught is mathematically represented as

$z_2 = \frac{x_2 - \mu_2}{\sigma_2 }$

substituting value

$z_2 = \frac{6 - 4}{0.8 }$

$z_2 = 2.5$

From our calculation we see that $z_2 > z_1$

The fish that was the better catch is the bass fish

3 0

3 years ago