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3 years ago

Pleas help me solve and can you show me how you solved it

Mathematics

2 answers:

NARA [144]3 years ago

6 0

Answer:

-8k + 6 = -10k + 10

k = 2

Step-by-step explanation:

Step 1: Add 10k to both sides.

−8k+6+10k=−10k+10+10k

2k+6=10

Step 2: Subtract 6 from both sides.

2k+6−6=10−6

2k=4

Step 3: Divide both sides by 2.

2k divided by 2 = 4 divided by 2

k=2

matrenka [14]3 years ago

3 0

Answer: K=2

Step-by-step explanation:

-8k+6=-10k+10

+10k

2k+6=10

-6

2k=4

k=2

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It would be approximately a decrease of 38%

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3 years ago

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3 years ago

Read 2 more answers

Oh and alsoo can someone plzzzzz add an explanation

Pani-rosa [81]

Answer:

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Step-by-step explanation:

42/6 = 7

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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:

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

And $P(X_i) \geq 0, \forall x_i$

So then we have a probability distribution

We can calculate the expected value with the following formula:

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

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:

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

6 0

4 years ago

Use the unit circle to find all values of thetabetween 0 and 2Ï€ for which sin(theta)=sqrt 3/2

ElenaW [278]

Sin(theta)=√3/2, 0≤theta≤360
In a unit circle, sin is y/r, r is always positive, y is positive in the first and second quadrant only.
sin60=√3/2, sin120=√3/2
so theta is 60 or 120

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3 years ago