Please help me answer this!!!

-19-5/4v is less than or equal too -14

Nesterboy [21]

Answer:

−20.25

Step-by-step explanation:

5 0

2 years ago

A local grocery story has strawberries on sale for $1.45 per pound. How much would it cost to buy 5.5 pounds of strawberries? Ro

myrzilka [38]

Answer: The cost of 5.5 pounds strawberries =$ 7.98

Step-by-step explanation:

Given: A local grocery story has strawberries on sale for $1.45 per pound.

i.e. Cost for 1 pound strawberries = $1.45

Now, Cost of 5.5 pounds strawberries = 5.5 x (Cost for 1 pound strawberries)

= $ (5.5 x 1.45)

= $ 7.975 ≈ $ 7.98 [ Round to the nearest penny]

Hence, the cost of 5.5 pounds strawberries =$ 7.98

5 0

3 years ago

Please help, thanks! Ok so, you pick a card at random, put it back, and then pick another card at random. There are FIVE cards a

tatyana61 [14]

Answer:

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

Step-by-step explanation:

Given

$S = \{1,2,3,4,5\}$

$n(S) = 5$

Required

$P(x > 1\ and\ x < 2)$

$P(x > 1\ and\ x < 2)$ is calculated as:

$P(x > 1\ and\ x < 2) = P(x > 1) * P(x < 2)$

Since it is a probability with replacement, we have:

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

For x > 1, we have:

$x > 1 = \{2,3,4,5\}\\$

$n(x > 1) = 4$

For x < 2, we have:

$x < 2 = \{1\}$

$n(x < 2) = 1$

$P(x > 1\ and\ x < 2) = \frac{n(x > 1)}{n(S)} * \frac{n(x < 2)}{n(S)}$

becomes

$P(x > 1\ and\ x < 2) = \frac{4}{5} * \frac{1}{5}$

$P(x > 1\ and\ x < 2) = \frac{4}{25}$

6 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

3 years ago

A dog walker charges a flat rate of $6 per walk plus an hourly rate of $30. How much does the dog walker charge for a 45 minute

diamong [38]

Answer:

- False

- True

- false

Step-by-step explanation:

- The variable that changes is the price

- 30 is the slope and 6 is the y-int

- 30(.75) + 6 = 28.5

5 0

3 years ago