Ask question

Ask question

All categories

3 years ago

15

Kyle wants to earn as many points as possible in one turn in a game. Two number cubes whose sides are numbered 1 through 6 are r

olled. He is given two options for the manner in which points are awarded in the turn. OPTION A: If the sum of the rolls is a prime number, Kyle receives 15 points. OPTION B: If the sum of the rolls is a multiple of 6, Kyle receives 42 points. Which statement best explains the option he should choose?

1 answer:

qwelly [4]3 years ago

5 0

Answer:

Kyle should choose Option B. The mathematical expectation of this option is 7 and is the greater mathematical expectation of the two options.

Step-by-step explanation:

i took the k-12 quiz

goodluck

You might be interested in

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

Randy and Sandy have a total of 20 books between them. After Sandy loses three by leaving them on the bus, and some birthday gif

nasty-shy [4]

Answer:

Randy had 13 books an Sandy had 7 books.

Step-by-step explanation:

20 - 3 = 17

30 - 17 = 13

13 x 2 = 26

26 + 4 = 30

13 + 7 = 20

Hope this helps! Please mark as brainliest.

3 0

3 years ago

According to a marketing research study, American teenagers watched 14.8 hours of social media posts per month last year, on ave

ki77a [65]

Answer:

The value of test statistics is 1.06.

Step-by-step explanation:

We are given that according to a marketing research study, American teenagers watched 14.8 hours of social media posts per month last year, on average. A random sample of 11 American teenagers was surveyed and the mean amount of time per month each teenager watched social media posts was 15.6. This data has a sample standard deviation of 2.5.

We have to test if the mean amount of time American teenagers watch social media posts per month is greater than the mean amount of time last year or not.

Let, NULL HYPOTHESIS, $H_0$ : $\mu$ = 14.8 hours {means that the mean amount of time American teenagers watch social media posts per month is same as the mean amount of time last year}

ALTERNATE HYPOTHESIS, $H_1$ : $\mu$ > 14.8 hours {means that the mean amount of time American teenagers watch social media posts per month is greater than the mean amount of time last year}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of time per month each teenager watched social media posts = 15.6 hours

$s$ = sample standard deviation = 2.5 hours

n = sample of teenagers = 11

So, <u>test statistics</u> = $\frac{15.6 - 14.8}{\frac{2.5}{\sqrt{11} } }$ ~ $t_1_0$

= 1.06

Hence, the value of test statistics is 1.06.

5 0

3 years ago

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago

What are examples of how statistics can lend credibility to an argument

lions [1.4K]

They show facts to prove that your opinion is right.

3 0

3 years ago

Read 2 more answers