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Stels [109]
3 years ago
5

Together, Victor and Tami Vargas earn $33,280 per year. Tami earns $4,160 more per year than Victor earns. How much do Victor an

d Tami each earn per year?
Mathematics
1 answer:
77julia77 [94]3 years ago
4 0

Answer: tami earns 20800

Victor earns 12480

Step-by-step explanation:

33280/2=16640

16640-4160=12480(victor)

16640+4160(tami)

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William Right broke his leg in an accidental fall. He has a deductible of $150 and a coinsurance payment equal to 20% percent of
Varvara68 [4.7K]

We know that William paid his deductible that amount is the $150. 

Then William paid 20 percent of everything else. That's 20 percent of (750 minus 150), it is equal to 150.

Then the insurance company paid 80 percent of (750 minus 250), it is equal to 480.

7 0
3 years ago
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Angle = (2x + 10)° and angle = (4y – 30)°. Find x and y.
ASHA 777 [7]

Answer:

11.33 degrees

Step-by-step explanation:

5 0
2 years ago
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}

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\lambda \geq 6.63835

4 0
3 years ago
What are the zeros of the quadratic function f(x) = 6x^2 + 12x - 7?
Sergeeva-Olga [200]

Answer:

-12 + √(144- 4(-42)/12

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Morgarella [4.7K]

Answer:  The answer is (C) 3 4 × 1 6 = 3 24


Step-by-step explanation:  We are given a rectangular picture made of square boxes and we are to find among the given options which is modelled by the given picture.

In the diagram, there are 6 boxes lengthwise and 4 boxes breadthwise. Also, 3 × 4 boxes are painted yellow and 1 × 6 boxes are painted blue. The common boxes will be 3.

Therefore, the correct choice is  3 4 × 1 6 = 3 24

Hence, (C) is the correct option.



3 0
4 years ago
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