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postnew [5]
3 years ago
14

The doctor has told Cal O'Ree that during his ten weeks of working out at the gym, he can expect each week's weight loss to be $

1\%$ of his weight at the end of the previous week. His weight at the beginning of the workouts is $244$ pounds. How many pounds does he expect to weigh at the end of the ten weeks?
Mathematics
1 answer:
Sergio [31]3 years ago
6 0

Answer:

220.67 pounds

Step-by-step explanation:

This problem can be treated as a compounded interest with a weekly rate of -1%. Cal's weight after 'n' weeks at a 1% loss rate per week is described by the following expression:

W = 244*(1-0.01)^n

For n = 10 weeks, Cal's weight is:

W = 244*(1-0.01)^{10}\\W=220.67\ pounds

He expects to weigh 220.67 pounds at the end of the ten weeks.

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Solve the system of equations.<br> 2.5y + 3x = 27<br> 5x – 2.5y = 5
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3 years ago
Each morning, Hungry Harry eats some eggs. On any given morning, the number of eggs he eats is equally likely to be 1, 2, 3, 4,
mamaluj [8]

Answer:

Mean = 35

Variance = 291.7

Step-by-step explanation:

Data provided in the question:

X : 1, 2, 3, 4, 5, 6

All the data are independent

Thus,

The mean for this case will be given as:

Mean, E[X] = \frac{\textup{Sum of all the observations}}{\textup{Total number of observations}}

or

 E[X] = \frac{\textup{1+2+3+4+5+6}}{\textup{6}}

or

E[X] = 3.5

For 10 days, Mean = 3.5 × 10 = 35

And,

variance = E[X²] - ( E[X] )²

Now, for this case of independent value,

E[X²] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{\textup{6}}

or

E[X²] = \frac{1+4+9+16+25+36}{\textup{6}}

or

E[X²] = \frac{91}{\textup{6}}

or

E[X²] = 15.167

Therefore,

variance = E[X²] - ( E[X] )²

or

variance = 15.167 - 3.5²

or

Variance = 2.917

For 10 days = Variance × Days²

= 2.917 × 10²

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3 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!
Dmitriy789 [7]

Answer:

7.11\leq-7.1

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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