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2 years ago

Push technology is also known as _____.

Mathematics

2 answers:

Zigmanuir [339]2 years ago

4 0

WEBCASTING

According to wikipedia ,

push technology is is a style of Internet-based communication where the request for a given transaction is initiated by the publisher or central server.

A webcast is a media presentation distributed over the Internet using streaming media technology to distribute a single content source to many simultaneous listeners/viewers.

never [62]2 years ago

3 0

Push technology is also know as Webcasting

You might be interested in

If a random variable X is exponentially distributed with parameter λ=2, then P(X≥1) is equal to

Rasek [7]

Answer:

D. 0.1353

B. 0.0473

Step-by-step explanation:

For an exponentially distributed random variable, the cumulative distribution function is:

$P(x\leq a) = 1 - e^{-\lambda a}\\P(x\geq a) = e^{-\lambda a}$

with parameter λ=2, then P(X≥1) is equal to:

$P(x\geq 1) = e^{-2*1}=0.1353$

D. 0.1353

with parameter λ=1.5, then P(2≤X≤4) is equal to

$P(2\leq x \leq 4) = P(x \leq 4) - P(x\leq 2)\\P(2\leq x \leq 4) =1 - (e^{-1.5*4}) - (1-(e^{-1.5*2}))\\P(2\leq x \leq 4) = 0.0473$

B. 0.0473

7 0

2 years ago

WILL BE BRANLIEST!!!

stiks02 [169]

Answer:

See explanation

Step-by-step explanation:

A. Solution set is in the attachment (dark green shading in the middle)

B. Yes, 2(5) + 3(1) ≤ 15; 13 is less than or equal to 15

5 + 1 > 3, 6 is greater than 3

C. (4, 2) means that he buys 4 sandwiches and 2 hot lunches

8 0

2 years ago

Read 2 more answers

Each inch on a map of Roswell, New Mexico represents $0.51\text{ mile}$. If two hills in Roswell are $7$ inches apart on the map

pav-90 [236]

Answer:

$3.57$

Step-by-step explanation:

we know that

The scale factor on a map is equal to

$\frac{1}{0.51}\frac{inch}{miles}$

by proportion

Find how many miles are in the real life $7$ inches apart on the map

$\frac{1}{0.51}\frac{inch}{miles}=\frac{7}{x}\frac{inches}{miles}\\ \\x=0.51*7\\ \\x=3.57 \ miles$

5 0

2 years ago

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function

vitfil [10]

Answer:

V(t) = 25000 * (0.815)^t

The depreciation from year 3 to year 4 was $2503.71

Step-by-step explanation:

We can model V(t) as an exponencial function:

V(t) = Vo * (1+r)^t

Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.

We have that Vo = 25000, r = -18.5% = -0.185, so:

V(t) = 25000 * (1-0.185)^t

V(t) = 25000 * (0.815)^t

In year 3, we have:

V(3) = 25000 * (0.815)^3 = 13533.58

In year 4, we have:

V(4) = 25000 * (0.815)^4 = 11029.87

The depreciation from year 3 to year 4 was:

V(3) - V(4) = 13533.58 - 11029.87 = $2503.71

7 0

2 years ago

The weight of people in a small town in missouri is known to be normally distributed with a mean of 154 pounds and a standard de

ad-work [718]

If a random sample of 20 persons weighed 3,460, the sample mean x-bar would be 3460/20 = 173 pounds.
The z-score for 173 pounds is given by:
$z=\frac{173-154}{29}=0.655$
Referring to a standard normal distribution table, and using z = 0.66, we find:
$P(\bar x\ \textless \ 173)=0.7454$
Therefore
$P(\bar x\ \textgreater \ 173)=1-0.7454=0.2546$
The answer is: 0.2546

4 0

2 years ago