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Pepsi [2]
3 years ago
10

Please help with 2 questions thank you!

Mathematics
2 answers:
IRISSAK [1]3 years ago
7 0
Question 1:
Match the order of the letters.
M matches with P, but N does not match with R

Thus, MN≠PR.
So, choice 1 is your answer.

Question 2:
Matching the letters, we know RT = NQ

7x-5 = 5x+11
Add both sides by 5
7x = 5x + 16
Subtract both sides by 5x
2x = 16
Divide both sides by 2
x = 8
Wait! We're looking for the length, not the value of x.
Plug in x = 8 into any of the two equations.

You get 51.
Your answer is the second choice.

Have an awesome day! :)
aleksandr82 [10.1K]3 years ago
4 0
1 is a, since the MN would match up with PQ and PR would match up with MO
2  7x - 5 = 5x + 11
2x - 5 = 11
2x = 16
x = 8

Plug x back in
7(8) - 5 = 56 - 5 = 51 so B
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3 years ago
The sum of three numbers is 36. The first number is two times the second number, and the second number is
MakcuM [25]

Answer:

o (16, 8, 12)

Step-by-step explanation:

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7 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
Can someone please help me with this
lana [24]

Answer:

3x-10+70°=180°(Supplementary angles)

3x+60°=180°

3x=180°-60°

3x=120°

3x/3 =120°/3

x=40°

Step-by-step explanation:

Hope that this is helpful.

Have a wonderful day.

6 0
3 years ago
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