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3 years ago

If w= -1, x =6 and y= -3, find the following 3w² - 5w + 1

Mathematics

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Step-by-step explanation:

3(1)² - 5(1) + 1

3x3 = 9

9-5+1

9-6 = 3

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Please help..I dont get this and please Dont put a Link below please.

lianna [129]

Answer:

x = 27, y = 74

Step-by-step explanation:

Hi again :)

Since the angles of 5x + 4 and the one adjacent to x = 14 are corresponding angles, they are equal. That means that (5x + 4) + (x + 14) = 180 degrees.

5x + 4 + x + 14 = 180

6x + 18 = 180

6x = 162

x = 27

Also, the angle (5x + 4) and (2y - 9) are vertical angles, they are equal in value. We know the value of x now, so we'll substitute that in and solve for y.

5(27) + 4 = 2y - 9

135 + 4 = 2y - 9

139 = 2y - 9

148 = 2y

y = 74

As always, lmk if you have questions.

3 0

3 years ago

. 7/8 = /48<br> A. 6<br> B. 1<br> C. 13<br> D. 42

hodyreva [135]

7/8=x/48

x=(7/8)48
x=(7)(48/8)
x=(7)(6)
x=42
D.42

4 0

3 years ago

Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat

drek231 [11]

Answer:

$E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5$

We can find the second moment given by:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496$

And we can calculate the variance with this formula:

$Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246$

And the deviation is:

$Sd(X) = \sqrt{1.246}= 1.116$

Step-by-step explanation:

For this case we have the following probability distribution given:

X 0 1 2 3 4 5

P(X) 0.031 0.156 0.313 0.313 0.156 0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

We can verify that:

$\sum_{i=1}^n P(X_i) = 1$