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VikaD [51]
3 years ago
9

What is the value of x? Enter your answer in the box. x= __

Mathematics
1 answer:
RoseWind [281]3 years ago
8 0

Answer:

x =20

Step-by-step explanation:

We can use similar triangles and proportions to solve.  Take the smaller triangle side on top and the larger triangle side on the bottom.

11                 10

---------- = --------------

11+121             10 +5x+10

Combine like terms

11                 10

---------- = --------------

132             5x+20

Using cross products

11*(5x+20) = 10*132

55x+220 = 1320

Subtract 220 from each side

55x+220-220 = 1320-220

55x = 1100

Divide by 55

55x/55= 1100/55

x =20

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Answer:

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Step-by-step explanation:

8 0
3 years ago
Which of the statements are true about the function f given by f(x) = 100-e? Select all that apply.
tatiyna

From the above function, it is clear that the value of f is never 0. Hence the statement that is true is (Option E), See explanation of same below.

<h3>What is the explanation for the above function?</h3>

Note that the function is related to Euler's number which is depicted as:
e ≈ 2.7182. The function is given as:

f(x) = 100 * e^{-x}

Assuming x = -2, we'd have:

100 * 2.7182^{-2}

= 271.82^{-2}

= 0.00001353354

Hence, even when x tends < 0 the function f(x) thus, is never 0. See the attached graph for confirmation.

Learn more about functions at:

brainly.com/question/25638609

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6 0
2 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
3 years ago
Plz plz plz plz help plz no LINKS
Viefleur [7K]
(2.7, 2.3)
(2, 1.7)
(1.7, 1.3)
again may be wrong
5 0
3 years ago
What is the reminder when (x^4+36 ) is divided by (x^2-8)
iragen [17]

Answer:  35

Step-by-step explanation:

3 0
3 years ago
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