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Morgarella [4.7K]
3 years ago
8

Can I get help on this problem showing my work?

Mathematics
2 answers:
Evgen [1.6K]3 years ago
6 0
cost \: of \: 1 \:bracelet \\(no\:shipping): \\72 \: dollars - 9 \: dollars \: shipping = \\ 63 \: \div 9 \: = cost \: of \: 1 \: bracelet \: \\ = 7 \: dollars
\\cost \: of \: shipping: \\ 9 \: dollars \: \div 9 \: bracelets \: = \\ 1 \: dollar\: per\:bracelet \\ \\ formula \: for \: cost \: overall: \\ 7y \: + y

y = quantity, for example:

I want to buy 1 bracelet =
1×7 = 7 + 1 = 8$ per bracelet
Anon25 [30]3 years ago
4 0

9x+9+72

9x= total cost

9= shipping fee

72= what you paid in all  

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A man drove 210 miles in 3.5 hours. If the posted speed limit was 55 mph (miles per hour) the entire way, approximately how many
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5mph over the speed limit

Step-by-step explanation:

In 1 hour the man drove 210/3.5=60 miles. 60mph is over the speed limit (55) by 5 mph.

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Which values of a and b make the following equation true?
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What value of X makes the equation -5x-(-7-4x)=-2(3x-4) true?
rjkz [21]

Answer: x= \frac{1}{5} Makes It True

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If I Helped, Please Mark Me As Brainliest, Have A Great Day :D

3 0
3 years ago
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Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie
ludmilkaskok [199]

Answer:

\lambda \geq 6.63835

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that X \sim Poisson(\lambda)

The probability mass function for the random variable is given by:

f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...

And f(x)=0 for other case.

For this distribution the expected value is the same parameter \lambda

E(X)=\mu =\lambda

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

P(X\geq 2)=1-P(X

Using the pmf we can find the individual probabilities like this:

P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}

P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}

And replacing we have this:

P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)

And we want this probability that at least of 99%, so we can set upt the following inequality:

P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99

And now we can solve for \lambda

0.01 \geq e^{-\lambda}(1+\lambda)

Applying natural log on both sides we have:

ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)

ln(0.01) \geq -\lambda+ln(1+\lambda)

\lambda-ln(1+\lambda)+ln(0.01) \geq 0

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}

Where :

f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)

f'(x_n)=1-\frac{1}{1+\lambda}

Iterating as shown on the figure attached we find a final solution given by:

\lambda \geq 6.63835

4 0
3 years ago
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