Find the value of x.

The total cost of Anja’s trip to the dentist was $628.35. She paid a flat fee of $89.95 which included the checkup and cleaning

Nuetrik [128]

I may be wrong but I am sure this is the equation...
$628.35= $89.95+4x

4 0

3 years ago

Simplify the expression. 4x - 3 + 8x - 5 - 3x

kow [346]

Answer:

9x-8

Step-by-step explanation:

4x-3+8x-5-3x

12x-3-5-3x

12x-8-3x

9x-8

6 0

3 years ago

Suppose you buy a CD for $500 that earns 2.5% APR and is compounded quarterly. The CD matures in 3 years. How much will this CD

Ilya [14]

We're going to use the compounded intrest formula:

$A = P(1+\frac{r}{n})^{nt}$

Where P is the initial cost
r is the rate as a decimal
n is the amount per year that you invest the rate
t is the amount of time at which you're checking how much it's worth (yrs)

Using this information, we can use:

$A = 500(1+\frac{0.025}{4})^{3*4} \approx 538.82$

So your answer will be B.

5 0

3 years ago

Read 2 more answers

When network cards are communicating, bits can occasionally be corrupted in transmission. Engineers have de- termined that the n

ycow [4]

Answer:

a) 11.40% probability of 5 bits being in error during the transmission of 1 kb

b) 11.60% probability of 8 bits being in error during the transmission of 2 kb

c) 0.01% probability of no error bits in 3kb

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the giveninterval.

Poisson distribution with mean of 3.2 bits/kb (per kilobyte).

This means that $\mu = 3.2kb$ , in which kb is the number of kilobytes.

(a) What is the probability of 5 bits being in error during the transmission of 1 kb?

This is P(X = 5) when $\mu = 3.2$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140$

11.40% probability of 5 bits being in error during the transmission of 1 kb

(b) What is the probability of 8 bits being in error during the transmission of 2 kb?

This is P(X = 8) when $\mu = 2*3.2 = 6.4$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160$

11.60% probability of 8 bits being in error during the transmission of 2 kb

(c) What is the probability of no error bits in 3kb?

This is P(X = 0) when $\mu = 3*3.2 = 9.6$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001$

0.01% probability of no error bits in 3kb

7 0

3 years ago

I solve kinda a bit but its hard (Algebra 2)

melisa1 [442]

Answer:

x=9

Step-by-step explanation:

The goal is to combine our knowledge of geometry area and algebra skills to solve this problems.

We know area of rectangle is

$l \times w$

where l is the length and w is the width. We know that one side is 3 less than the other side. We can use L or W but let use L. Let x replace L

One side is x, while the other side is

$x - 3$

Both multiply to 54.

$x(x - 3) = 54$

${x}^{2} - 3x = 54$

${x}^{2} - 3x - 54 = 0$

$(x - 9)(x + 6)$

$x = 9$

or

$x = - 6$

Distance. is positive so x=9

8 0

3 years ago