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Sav [38]
4 years ago
8

Use the normal distribution of fish lengths for which the mean is 1111 inches and the standard deviation is 44 inches. assume th

e variable x is normally distributed. left parenthesis a right parenthesis(a) what percent of the fish are longer than 1212 ​inches? left parenthesis b right parenthesis(b) if 300300 fish are randomly​ selected, about how many would you expect to be shorter than 77 ​inches?

Mathematics
1 answer:
melomori [17]4 years ago
5 0
(a) P(x > 12) ≈ 0.4013
  About 40% of the fish are longer than 12 inches.

(b) 300*P(x < 7) ≈ 48
  about 48 of 300 fish could be expected to be shorter than 7 inches.

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Answer:

a) <em>The equation</em> (10s + 8w) <em>represents </em><em>the </em><em>calories </em><em>Bridget </em><em>ate </em><em>on </em><em>Monday </em><em>and </em><em>the </em><em>equation</em> (20s + w) <em>represents </em><em>the </em><em>calories</em><em> </em><em>she </em><em>ate</em><em> </em><em>the </em><em>next </em><em>day.</em>

<em>b)</em><em> </em><em>The </em><em>number </em><em>of </em><em>calories</em><em> </em><em>in </em><em>each </em><em>strawberry</em><em> </em><em>is </em>4 <em>and </em><em>the </em><em>number </em><em>of </em><em>calories </em><em>in </em><em>each </em><em>vanilla</em><em> </em><em>wafer</em><em> cookie</em><em> </em><em>is </em>19. The solution is s= 4 and w = 19.

Step-by-step explanation:

For part A, Bridget ate 10 strawberries and 8 vanilla wafer cookies on Monday. Since the the number of calories in a strawberry is <em>s</em> and the number of calories in a vanilla wafer cookie is <em>w </em>, the number of calories Bridget ate on Monday is <em>10s + 8w</em><em>.</em><em> </em>The next day, Bridget ate 20 strawberries and 1 vanilla wafer cookie. Hence, the number of calories Bridget ate on the next day is 20s<em> + w</em>.

For part B,

we will create two different simultaneous equations.

Equation 1: 10s + 8w = 192

Equation 2: 20s + w = 99

We need to find one of the terms first to solve the other term. For this case, I will solve for w first.

Multiply the first equation by 2.

Equation 3: 20s + 16w = 192*2 = 384.

Now, subtract equation 2 from this new equation.

Equation 4:

(20s + 16w) - (20s + w) = 384 - 99

20s + 16w - 20s - w = 285

15w = 285

This leaves only w left and we can solve w.

w = 285 / 15 = 19

Now, we can solve for s using equation 2.

20s + 19 = 99

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Hence,

s = 80/20 = 4

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