How do you do this problem?

write the product of the expression 5^4 * 5^-7 using a positive exponent, then write the product using a negative exponent

pav-90 [236]

Answer with positive exponent = $\left(\frac{1}{5}\right)^3$ or $\frac{1}{5^3}$

Answer with negative exponent = $5^{-3}$

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Explanation:

The rule we use is

$a^b*a^c = a^{b+c}$

If we multiply two exponential expressions with the same base, then we add the exponents.

The base for each is 5. The exponents 4 and -7 add to -3.

This means

$5^4*5^{-7} = 5^{4+(-7)} = 5^{-3}$

To convert to a positive exponent, we apply the reciprocal to the base. We go from 5, aka 5/1, to 1/5.

So, $5^{-3} = \left(\frac{1}{5}\right)^3 = \frac{1^3}{5^3} = \frac{1}{5^3}$

3 0

3 years ago

What is the slope of the line?

vivado [14]

Do rise over run and you should end up with 4 and 3 hope that help

3 0

3 years ago

What is equal to 19^-8?

Rudiy27

Answer:

5.888045975

Step-by-step explanation:

4 0

3 years ago

The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associate

Schach [20]

Answer:

(a)0.16

(b)0.588

(c) $[s_1$ s_2]=[0.75,$ 0.25]$

Step-by-step explanation:

The matrix below shows the transition probabilities of the state of the system.

$\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix $P^k$ .

(a)

$P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

$P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}$

If the system is initially running, the probability of the system being down in the next hour of operation is the $(a_{12})th$ entry of the P^2$ matrix.$

The probability of the system being down in the next hour of operation = 0.16

(b)After two(periods) hours, the transition matrix is:

$P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}$

Therefore, the probability that a system initially in the down-state is running

is 0.588.

(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.

Since we have two states, $S=[s_1$ s_2]$

$[s_1$ s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$ s_2]$

Using a calculator to raise matrix P to large numbers, we find that the value of $P^k$ approaches [0.75 0.25]:

Furthermore,

$[0.75$ 0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75$ 0.25]$

The steady-state probabilities of the system being in the running state and in the down-state is therefore:

$[s_1$ s_2]=[0.75$ 0.25]$

4 0

4 years ago

Stella buys 6 tacos and 4 enchiladas for 52 dollars. Eliana buys 7 tacos and 9 enchilladas for 78 dollars.

Firdavs [7]

Answer:

Let tacos be x and enchiladas be y

6 tacos + 4 enchiladas = 52

becomes

6x+4y=52

^^that's our first equation

Now, the second equation

7 tacos + 9 enchiladas = 78

7x+9y = 78

^^^that's our second equation

Isolate y in the first equation:

6x+4y=52

4y=52-6x

y = (52-6x)/4

^^^Let's call this equation 3

Now, sub-in equation 3 into equation 2

*replace the y*

Equation 2: 7x+9y = 78

Sub-in

7x+ <u>9 ( (52-6x)/4)</u> = 78

Do some algebra (get your x terms on one side) and you will find your x value

7x + <u>117 - 13.5x</u> = 78

(7x-13.5x) = 78-117 *get your x terms on one side

x=6

That's the cost of one taco.

Now, find y. Pick an equation (1,2 or 3). Let's pick 3 since y is already isolated.

All you do is plug in your x value.

Equation 3: y = (52-6x)/4

y= (52 -6(6))/4

...

You should be able to solve this on your own

Step-by-step explanation:

:)) You can solve for y now

5 0

2 years ago