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Tomtit [17]
2 years ago
14

The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and

changed the amount of food he was feeding him. Rocco lost half a pound the first week. Jake wants to determine Rocco’s weight in pounds, p, after w weeks if Rocco continues to lose weight based on his vet’s advice.
Mathematics
2 answers:
madreJ [45]2 years ago
6 0

Answer:it would take him five weeks to lose ten pounds

Step-by-step explanation:

Lesechka [4]2 years ago
5 0

Answer: To determine Rocco´s weight jake must use this expression

  • p = 55 - 0.5 w

Step-by-step explanation:

Hi to solve this problem we have to analyze the information given:

  • Rocco´s initial weight: 55 pounds
  • Pounds lost per week: 0.5 pounds  

So, we have to write an equation where p is Rocco's weight after w weeks.

Each week Rocco will lose 0.5 pounds so the expression is 0.5w, where w is the number of weeks.

Mathematically speaking:

  • p = 55 - 0.5 w

By replacing w by the number of weeks, Jake will obtain Rocco´s weight.

For example, if the number of weeks is 20:

p= 55-0.5x 10

p= 55 - 20

p= 35

After 20 weeks Rocco will lose 10 pounds.

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A water mill grinds corn to make cornmeal. The water wheel has a radius of 18 feet. The wheel is rotating at 8 revolutions per m
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Answer:

904.9 feet per minute

Step-by-step explanation:

The Angular distance can be calculated using below expression

S= r θ ..............eqn(1)

Where

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But angular distance can also be expressed as

θ = wt...................eqn(2)

w= angular speed= 8 rev per minute= (8 × 2π)

t= time

Velocity can also be expressed as

V= s/t............eqn(3)

Where

V= velocity

s= distance

t= time

Substitute eqn(2) into eqn(3)

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Substitute eqn(2) into eqn(4)

V=( r w t)/ t

V= r w ..............eqn(5)

V= 18×(8 ×2π)

=288π

= 904.9 feet per minute

Hence, the linear speed, in feet per minute, of the water. is 904.9 feet per minute

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2 years ago
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A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
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Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0
2 years ago
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