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alukav5142 [94]

2 years ago

5

John is playing a board game where all the players pay him if he lands on a certain space. When he lands on that space, he has t

o roll two six-sided dice and add them together. That is the amount each player has to pay him. John lands on that space and then rolls a 6 and a 5. If he is playing with 4 other players and already has $123 , how much will he have after each player pays him?

1 answer:

Zepler [3.9K]2 years ago

7 0

Answer:

The answer is $523

Step-by-step explanation:

Because well think about it at first he has $123 and there r four players and so they each give $100 every time he wins so that will be $523 that he will have after.

You might be interested in

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago

Why can i keep giving brainlyest? they should make it like every day you can give 1

defon

That’s true!!!! They totally should do that!

6 0

3 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

3 0

3 years ago

Solve the equation and graph the solution(s), if possible. 9|4p + 2| + 8 = 35

Cerrena [4.2K]

Solving an equation means finding a value of a variable that make the equation true. This value is called the solution of the equation.<span> A mathematical equation is just like a balanced equation. Both sides of the equation should be equal at all times. </span><span>Whatever we do to one side, has to be done on the opposite side.</span>

$9|4p + 2| + 8 = 35$

$
9|4p+2|=35-8
 
$ $Subtract \ 8 \ from \ both \ sides$

$9|4p+2|=27

 
$

$|4p+2|= \dfrac{27}{9}$ $Divide \ both \ sides \ by \ 9$

$|4p+2|=3$

$Break \ down \ the \ problem \ into \ these \ 2 \ equations$

$
4p+2=3
 
$

$-(4p+2)=3$

$Solve \ the \ 1st \ equation: \ 4p+2=3$

$4p=3-2$   <span> $Subtract \ 2 \ from \ both \ sides$
</span>
$4p=1$

$p= \dfrac{1}{4}$   <span> $Divide \ both \ sides \ by \ 4$
</span>
$Solve \ the \ 2nd \ equation: \ -(4p+2)=3$

$-4p−2=3$    $Simplify \ brackets$

$-4p=3+2$ <span> $Add \ 2 \ to \ both \ sides$
</span>
$-4p=5$

$p= -\dfrac{5}{4}$

$Collect \ all \ solutions$

$p= -\dfrac{5}{4}$ , $p= \dfrac{1}{4}$

8 0

3 years ago

If takes 660 metres of fence to enclose a rectangular field. if one side is 30 metres less that twice the other side, find the d

aliina [53]

Answer:

210 by 120

Step-by-step explanation:

perimeter=2l+2w

length= 2w-30

width= w

2(2w-30+w) = 660

6w-60= 660

6w= 720

w=120

length= 2*120-30 = 210

6 0

2 years ago