If a 2:1 transformer has 220 vac input a voltage test at the primary winding should produce a meter reading of

Say a hacker has a list of n distinct password candidates, only one of which will successfully log her into a secure system. a.

alexdok [17]

Answer:

The probability is $\frac{1}{n}$

Step-by-step explanation:

If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:

If k=1

$P = \frac{1}{n}$

Because, in her first try she has n possibles options and just one give her a successful login.

If k=2

$P=\frac{n-1}{n} *\frac{1}{n-1} =\frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.

If k=3

$P=\frac{n-1}{n} *\frac{n-2}{n-1} *\frac{1}{n-2} = \frac{1}{n}$

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.

Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n

7 0

3 years ago

What inequality represents the following sentence?

Maurinko [17]

C.

Hope this helps! :)

7 0

3 years ago

Find the gradients of line a and b

gayaneshka [121]

Answer:

Gradient of A: 2

Gradient of B: -1

Step-by-step explanation:

Gradient = change in y/change in x

✔️Gradient of A using two points on line A, (2, 5) and (0, 1):

Gradient = (1 - 5)/(0 - 2) = -4/-2

Simplify

Gradient of A = 2

✔️Gradient of B using two points on line B, (0, 5) and (5, 0):

Gradient = (0 - 5)/(5 - 0) = -5/5

Simplify

Gradient of B = -1

5 0

3 years ago

Help me please I’ll mark brainliest thank you

jenyasd209 [6]

Answer:

11/15=x/240

240/15=16

11 x 16=176

11/15=176/240 or rather he completes 176 passes

Step-by-step explanation:

4 0

3 years ago

Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa

Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation

n=580 represent the random sample taken

X represent the seafood sold in the country that is mislabeled or misidentified by the people

$\hat p=0.25$ estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

$\alpha=0.01$ represent the significance level (no given, but is assumed)

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204$

$0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296$

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).

Part c

A government spokesperson claimed that the sample size was too small, relative to the billions of pieces of seafood sold each year, to generalize. Is this criticism valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0

4 years ago