What is the fraction of 10000000000

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed

algol [13]

Answer:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

$p_v = P(\chi^2_{5} >5.860)=0.32$

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Number: 1, 2 , 3 , 4 , 5 ,6

Frequency: 27, 31, 42, 40, 28, 32

We need to conduct a chi square test in order to check the following hypothesis:

H0: The outcomes are equally likely.

H1: The outcomes are not equally likely.

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are given:

$O_{1}=27$ $O_{2}=31$

$O_{3}=42$ $O_{4}=40$

$O_{5}=28$ $O_{6}=32$

The expected values are given by:

$E_{1} =\frac{1}{6}*200=33.33$ $E_{2} =\frac{1}{6}*200=33.33$

$E_{3} =\frac{1}{6}*200=33.33$ $E_{4} =\frac{1}{6}*200=33.33$

$E_{5} =\frac{1}{6}*200=33.33$ $E_{6} =\frac{1}{6}*200=33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >5.860)=0.32$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(5.860,5,TRUE)"

Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.

4 0

3 years ago

Elsie bought a video game for $85. If the store charges 170% of its cost for the game, how much is the store's profit? *Hint: On

nikklg [1K]

Answer:

$ 59.5

Step-by-step explanation:

Calculation for how much is the store's profit

First step is to calculate the store's cost

Store's cost= 170/100 * $85

Store's cost =$144.5

Now let calculate the store's profit

Store'sProfit=$144.5-$85

Store's Profit=$ 59.5

Therefore the store's profit will be $59.5

7 0

2 years ago

F(x) = e^x/2 Find F(0). 1.36 1 0

Lelu [443]

Answer:

f(0)

= e^o/2

= 1/2

=0.5 it is the required answer

8 0

3 years ago

Read 2 more answers

Classify the real numbers as rational or irrational numbers.

alex41 [277]

Answer:

Rational Numbers: 1.1625, 5.15603418923, and -11/151

Irrational Numbers: √256, √141, and ∛25

Step-by-step explanation:

Rational Numbers are numbers that are decimals, fractions, and whole numbers.

Irrational Numbers are numbers that have square roots and cube roots.

7 0

2 years ago

A study investigated about 3000 meals ordered from Chipotle restaurants using the online site Grubhub. Researchers calculated th

Allisa [31]

Answer:

21.186%

Step-by-step explanation:

z = (x-μ)/σ,

where

x is the raw score = 2400mg

μ is the population mean = 2000mg

σ is the population standard deviation = 500mg

z = 2400 - 2000/500

z = 0.8

Probability value from Z-Table:

P(x<2400) = 0.78814

P(x>2400) = 1 - P(x<2400) = 0.21186

Converting to percentage:

0.21186 × 100

= 21.186%

Therefore, the percent of the meals

ordered that exceeded the recommended daily allowance of 2400 mg of sodium is 21.186%

7 0

3 years ago