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2 years ago

6x^{2} + 19 + 9x^{2} - 8 With Explanation

Mathematics

1 answer:

NemiM [27]2 years ago

7 0

Answer:

I'mnot sure if you're asking for its roots..but here you go :)

Step-by-step explanation:

$= 6 {x}^{2} + 19 + 9 {x}^{2} - 8 \\ = (6 {x}^{2} + 9 {x}^{2} ) + (19 - 8) \\ = 15 {x}^{2} + 8 \\ 15 {x}^{2} = - 8 \\ {x}^{2} = - \frac{8}{15} \\ \sqrt{ {x}^{2} } = \sqrt{ - \frac{8}{15} } \\ x = \sqrt{ \frac{8}{15}} \times \sqrt{ - 1} \\ x = \sqrt{ \frac{8}{15} } \times i \\ x1 = 0.7302i \\ x2 = - 0.7302i$

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Confused can someone help

Alex

Answer:

B. 3x^2 - 2x - 1.

Step-by-step explanation:

(f + g)x = f(x) + g(x)

= -4x + 3 + 3x^2 + 2x - 4

Adding like terms we get the answer:

= 3x^2 - 2x - 1.

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3 years ago

Read 2 more answers

. Katie withdrew a total of $300 from her bank account over the last 5 weeks. How much money did she withdraw each week? Write a

Elodia [21]

Answer:

300/5=60

Step-by-step explanation:

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3 years ago

What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success p? C

charle [14.2K]

Answer:

$(D)E[ X ] =np.$

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success p,

$f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq x\leq n$

$E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}$

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

$E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}$

Now,

$x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)$

Substituting,

$E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}$

Factoring out the n and one p from the above expression:

$E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}$

Representing k=x-1 in the above gives us:

$E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}$

This can then be written by the Binomial Formula as:

$E[ X ] = (np) (p +(1 - p))^{n -1 }= np.$

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3 years ago

Solve the given system of equations.

Goryan [66]

The given equations are:
1) 2y = -x + 9
⇒ x = 9-2y

2) 3x - 6y = -15
⇒3x = 6y - 15
x = 2y - 5

Equating the values of x, we get:

9 - 2y = 2y - 5
9 + 5 = 4y
14 = 4y
y = 3.5

Using this value of y in equation 1 we get:

x = 9 - 2(3.5) = 2

So, the solution set is (2, 3.5)

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3 years ago

A grocery store sells salmon fillets in package that contains at least 3 fillets of salmon,but no more than 5.which of the follo

34kurt

Answer: 18 ≤ 6*S ≤ 30

Where S is the number of salmon fillets in one single package.

Step-by-step explanation:

Let's define S as the number of salmon fillets in one package.

We know that it contains at least 3, and no more than 5, then we can write:

3 ≤ S ≤ 5

Now we want to know the total number of salmon fillets that could be on 6 packages, then if S is the number of salmon fillets in one package, 6*S will e the number of salmon fillets in 6 packages.

We can find this by multiplying the inequality:

3 ≤ S ≤ 5

by 6.

We get:

6*3 ≤ 6*S ≤ 6*5

18 ≤ 6*S ≤ 30

Then in 6 packages, we could have any number between 18 and 30 fillets of salmon.

8 0

3 years ago