Ask question

Ask question

All categories

Free_Kalibri [48]

3 years ago

5

Find the value of a and z: x5⋅x4=axz a = , z = Find the values of a and z: 6x20+5x20=axz a = , z = Find the value of a and z: (x

5)4=axz a = , z = Find the value: x25−x25 = Find the values of a, b and z: 8x−9=abxz a = , b = , z =

1 answer:

djyliett [7]3 years ago

5 0

Z is equal to z because is z

You might be interested in

Write the equation of the line that passes through (0,3) and (-4,-1).

BlackZzzverrR [31]

Answer:

number 1

Step-by-step explanation:

if you'd look at Number One X is negative and if you look at the two number problems the x that is negative is negative for and then the one that is positive is the Y which is 3

3 0

3 years ago

What number should be placed in the box to help complete the division calculation?

Drupady [299]

Multiply 13 by the 2 above the divide line

13*2 = 26
so the number in the box should be 260

5 0

3 years ago

Evaluate the expression x = 4.<br> 3x + 2

Mekhanik [1.2K]

Answer:

14

Step-by-step explanation:

3(4)+2

12+2

14

7 0

3 years ago

Find the value <br> of x so that the quadrilateral is a parallelogram.

Lelu [443]

Answer:

46

Step-by-step explanation:

It must be the same as the angle opposite of it.

You can see that 134 and 134 are the same, so the other two corners need to be the same and be 46 and 46.

8 0

3 years ago

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

Licemer1 [7]

Answer:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

1) $f(x_i) \geq 0, \forall x_i$

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

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

Step-by-step explanation:

For this case we have the following density function:

$f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3$

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) $f(x_i) \geq 0, \forall x_i$

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

We can find the individual probabilities and we got:

$f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923$

$f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307$

$f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770$

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

$P(X \leq 1) = P(X=1) =0.6923$

$P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077$

$P(2$

7 0

2 years ago