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2 years ago

A computer randomly selects a letter from the alphabet.

Mathematics

1 answer:

Nikolay [14]2 years ago

7 0

1) 26 different outcomes are in the sample space.

2) 1 / 26 is the probability that the computer produces the first letter of your first name.

Step-by-step explanation:

1) You have to find out the different outcomes in the sample space :

A "Sample space" is defined as the set of all the possible outcomes of an event.
Here, the given event is randomly selecting a letter from the alphabets.

Therefore, the sample space must contain all the possible alphabets that can be chosen randomly.

The sample space is the set of all the 26 alphabets in English language.

⇒ Sample space = {A,B,C,D...........,Y,Z}

⇒ 26 different outcomes.

2) The probability the computer produces the first letter of your first name :

Here, the required outcome is getting the first letter of your first name.

Probability = No. of required outcomes / total no. of outcomes.

For example, The name Alex Davis has the first letter of the fist name as alphabet 'A'.

∴ Probability = 1 / 26

Similarly, for any first name there is going to be any one alphabet from the 26 alphabets, thus the probability to get the first letter will be always 1 / 26.

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What’s 4x4x2-4 in a fraction over 2

JulijaS [17]

Answer:

it should be 12/2 or 6 i believe

Step-by-step explanation:

4x4 = 8

8x2 = 16

16-4 = 12

8 0

2 years ago

What is 15 tens - 1 gross

vampirchik [111]

Answer:

see the explanation

Step-by-step explanation:

we know that

A gross is equal to 120 ones or ten dozen

what is 15 tens - 1 gross

we know that

15 tens means ----> That you are adding 10, 15 times or multiplying 10 by 15, which gives you

$10(15)=150$

1 gross means ---> That you are adding 10, 12 times or multiplying 10 by 12

which gives you

$10(12)=120$

The algebraic expression of 15 tens - 1 gross is equal to

$150-120=30$

Convert to word expression

3 tens

8 0

2 years ago

What number must we multiply by $-\frac23$ to get a product of $\frac34$?

stich3 [128]

Let x be unknown number. If number x is multiplied by $-\dfrac{2}{3}$ and the product is equal to $\dfrac{3}{4},$ then

$x\cdot \left(-\dfrac{2}{3}\right)=\dfrac{3}{4}.$

To find x you should divide $\dfrac{3}{4}$ by $-\dfrac{2}{3}:$

$x=\dfrac{\dfrac{3}{4}}{-\dfrac{2}{3}}=\dfrac{3}{4}\cdot \left(-\dfrac{3}{2}\right)=-\dfrac{3\cdot 3}{4\cdot 2}=-\dfrac{9}{8}.$

Answer: $x=-\dfrac{9}{8}$

7 0

2 years ago

The first term of a finite geometric series is 6 and the last term is 4374. The sum of all the term os 6558. find the common rat

Oliga [24]

A geometric series is written as $ar^n$ , where $a$ is the first term of the series and $r$ is the common ratio.

In other words, to compute the next term in the series you have to multiply the previous one by $r$ .

Since we know that the first time is 6 (but we don't know the common ratio), the first terms are

$6, 6r, 6r^2, 6r^3, 6r^4, 6r^5, \ldots$ .

Let's use the other information, since the last term is $4374 > 6$ , we know that $r>1$ , otherwise the terms would be bigger and bigger.

The information about the sum tells us that

$\displaystyle \sum_{i=0}^n 6r^i = 6\sum_{i=0}^n r^i = 6558$

We have a formula to compute the sum of the powers of a certain variable, namely

$\displaystyle \sum_{i=0}^n r^i = \cfrac{r^{n+1}-1}{r-1}$

So, the equation becomes

$6\cfrac{r^{n+1}-1}{r-1} = 6558$

The only integer solution to this expression is $n=6, r=3$ .

If you want to check the result, we have

$6+6*3+6*3^2+6*3^3+6*3^4+6*3^5+6*3^6 = 6558$

and the last term is

$6*3^6 = 4374$

7 0

2 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

2 years ago