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anzhelika [568]
2 years ago
9

A coin will be tossed 100 times. You get to pick 11 numbers. If the number of heads turns out to equal one of your 11 numbers, y

ou win a dollar.
Which 11 numbers should you pick, and what is your chance (approximately) of winning? Explain.
Mathematics
1 answer:
kati45 [8]2 years ago
8 0

Answer:

Numbers:

45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55

Chance:

68.27%

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Answer:

Slope: 2

Y-intercept: (0,-7)

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What is the mean number of hours worked that week?
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well how menny hours per day?

Step-by-step explanation:

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3 years ago
Evaluate the expression 40÷1+3-(3×7)+7-5
Eduardwww [97]
For this problem, you will have to use PEMDAS  (Parentheses) (Exponents) (Multiplication/Division) (Addition/Subtraction)

First of all, you will want to solve for the values inside the parentheses first

40 <span>÷ 1+3-(21)+7-5      Now check for anything that includes                                                                   multiplication/division and do that

40+3-21+7-5              All you have left now is addition and subtraction. It                                           must go from left to right, therefore if addition is first,                                       do addition, if subtraction is first, then do subtraction.

Final answer = 24</span>
8 0
3 years ago
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A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
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