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Andrej [43]
3 years ago
13

Can someone buy me eddie van der tabs???

Computers and Technology
1 answer:
sineoko [7]3 years ago
4 0

Answer:

No

Explanation:

U cannot exploit people like that

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Which app or website teaches java and phyton
aev [14]

Microsoft is a website that teaches java and phyton

6 0
3 years ago
Your boss bought a new printer with a USB 3.0 port, and it came with a USB 3.0 cable. Your boss asks you: Will the printer work
Greeley [361]

Answer:

Yes, is should work

Explanation:

USB is widely adopted and supports both forward and backward compatibility. The USB 3.0 printer should work with the USB 2.0 computer. However, having a connection like this, the printer will only be able to work at the speeds of the computer’s USB 2.0. By default, USB is built to allow transfer speeds improvement with upgrades from previous generations while still maintaining compatibility between devices that are supported by them.

4 0
3 years ago
The range of an area where users can access the Internet via high-frequency radio signals transmitting an Internet signal from a
Rama09 [41]

Answer:

A hot spot

Explanation:

Q:

The range of an area where users can access the Internet via high frequency radio signals transmitting an Internet signal from a wireless router is known as a _____. A) HotspotB) PAN…

A:

A) hotspot Bluetooth is for short distance and pan is Personal area networks (PANs) connect an individual's personal devices

4 0
2 years ago
The fastest way to get help is to type a word or two in the search box.
Elanso [62]
The fastest way to get help is to type a word or two in the search box. TRUE.
5 0
3 years ago
Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

Sample size(n) = 250

np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

6 0
3 years ago
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