Ejercicios de velocidad angular

Ill give brainlist to whoever helps me w this question

spayn [35]

The answer is C because multiplication comes first

7 0

2 years ago

g A psychic was tested for extrasensory perception (ESP). The psychic was presented with cards face down and asked to determine

diamong [38]

Answer:

Step-by-step explanation:

Hello!

The objective is to test ESP, for this, a psychic was presented with cards face down and asked to determine if each of the cards was one of four symbols: a star, cross, circle, square.

Be X: number of times the psychic identifies the symbols on the cards correctly is a size n sample.

p the probability that the psychic identified the symbol on the cards correctly

You have to calculate the sample size n to estimate the proportion with a confidence level of 95% and a margin of error of d=0.01

The CI for the population proportion is constructed "sample proportion" ± "margin of error" Symbolically:

p' ± $Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$

Where $d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$ is the margin of error. As you can see, the formula contains the sample proportion (it is normally symbolized p-hat, in this explanation I'll continue to symbolize it p'), you have to do the following consideration:

Every time the psychic has to identify a card he can make two choices:

"Success" he identifies the card correctly

"Failure" he does not identify the card correctly

If we assume that each symbol has the same probability of being chosen at random P(star)=P(cross)=P(circle)=P(square)= 1/4= 0.25

Let's say, for example, that the card has the star symbol.

The probability of identifying it correctly will be P(success)= P(star)= 1/4= 0.25

And the probability of not identifying it correctly will be P(failure)= P(cross) + P(circle) + P(square)= 1/4 + 1/4 + 1/4= 3/4= 0.75

So for this experiment, we'll assume the "worst case scenario" and use p'= 1/4 as the estimated probability of the psychic identifying the symbol on the card correctly.

The value of Z will be $Z_{1-\alpha /2}= Z_{0.975}= 1.96$

Now using the formula you have to clear the sample size:

$d= Z_{1-\alpha /2} * (\sqrt{\frac{p'(1-p')}{n} } )$

$\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p')}{n} }$

$(\frac{d}{Z_{1-\alpha /2}})^2 =\frac{p'(1-p')}{n}$

$n*(\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')$

$n = p'(1-p')*(\frac{Z_{1-\alpha /2}}{d})^2$

$n = (0.25*0.75)*(\frac{1.96}{0.01})^2= 7203$

To estimate p with a margin of error of 0.01 and a 95% confidence level you have to take a sample of 7203 cards.

I hope this helps!

5 0

3 years ago

Read 2 more answers

When a number is decreased by 40% of itself the result is 96. What is the number?

Nikitich [7]

Answer:

160

Step-by-step explanation:

96 / (100%-40%) = 96/ (60%)

= 96/0.6 = 160

3 0

2 years ago

Read 2 more answers

Write the given vector in terms of i and j.

8_murik_8 [283]

The first number is coefficient for i, and the second number is coeff. for j
=>
<0,-8> is the same as 0i-8j, or simply 8j

5 0

3 years ago

Current rules for telephone area codes allow the use of digits 2-9 for the first digit, and 0-9 for the second and third dig

Anna35 [415]

Using the fundamental counting theorem, we have that:

648 different area codes are possible with this rule.
There are 6,480,000,000 possible 10-digit phone numbers.
The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.

For the area code:

8 options for the first digit.
9 options for the second and third.

Thus:

$8 \times 9 \times 9 = 648$

648 different area codes are possible with this rule.

For the number of 10-digit phone numbers:

7 digits, each with 10 options.
648 different area codes.

Then

$648 \times 10^7 = 6,480,000,000
$

There are 6,480,000,000 possible 10-digit phone numbers.

The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.

A similar problem is given at brainly.com/question/24067651

5 0

2 years ago