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fredd [130]
2 years ago
12

1. What is the solution to the equation 3(x - 2)= 15?​

Mathematics
2 answers:
Dahasolnce [82]2 years ago
5 0

Answer:

x=7

Step-by-step explanation:

3x-6=15

3x-6+6=15+6

3x=15+6

3x = 21

----   ----

3       3

x=7

Neko [114]2 years ago
4 0

Answer:

X=7

Step-by-step explanation:

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Find the mean, median, and mode for 0, 4, 6, 11, 9, 8, 9, 1, 5, 9, 7. Round to the nearest tenth if needed.
snow_lady [41]

Answer:

6

Step-by-step explanation:

you add them all up and divide it by how many numbers there are

6 0
3 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}

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

\lambda \geq 6.63835

4 0
3 years ago
Find the mean absolute deviation for the data set.<br> 5,6,6,8,10
alukav5142 [94]

Answer: 1.6

<u>Order the numbers</u>

5,6,6,8,10

<u>Add</u>

5+6+6+8+10=35

<u>Divide</u>

35÷5=7

Mean: 7

Sum divided by the count.

Final Answer: 1.6

6 0
3 years ago
Read 2 more answers
4x+6*2x combine like terms
AleksAgata [21]

Answer:

16x

Step-by-step explanation:

4x + 12x = 16x

5 0
2 years ago
Read 2 more answers
The length of a rectangle is eight feet more than its width. The area of the rectangle is six hundred nine square feet. What are
vodomira [7]
You can write two equations using the given information:
.. L = W +8
.. L * W = 609
Using substitution, you get a quadratic.
.. (W +8) * W = 609
.. W^2 +8W -609 = 0
Not surprisingly, you're looking for factors of 609 that differ by 8.
.. 609 = 1*609 = 3*203 = 7*87 = 21*29
The last two are the factors of interest.
.. (W +29)(W -21) = 0

The width of the rectangle is 21 feet, the length is 29 feet.

_____
Sometimes it is easier to work with the average dimension. Here, let that be x. Then you have
.. (x +4)(x -4) = 609
.. x^2 -16 = 609
.. x^2 = 625 = 25^2 . . . . . . . one of your memorized math facts
So, the dimensions are
.. 25 +4 = 29 by 25 -4 = 21, that is, 29 ft by 21 ft.
5 0
3 years ago
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