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3 years ago

A 12-sided number cube with the numbers 1 through 12 is rolled.

2 answers:

8 0

Hello!

You are finding the probability that you will roll a number greater than 10

You do this by doing the amount of favorable outcomes over the total amount of outcomes

There are 2 numbers greater than 10

There are 12 total numbers

This is a probability of 2/12

2/12 = 0.17

The answer is A)0.17

Hope this helps!

weeeeeb [17]3 years ago

4 0

Total number of possible outcomes = 12
Total number of favourable outcomes (i.e. greater than 10) = 2

P(greater than 10) = 2/12 = 0.17

Answer: 0.17

You might be interested in

What is the rational number that is equivalent to a -24 / -6

saw5 [17]

Answer:

Step-by-step explanation:

hi! let's divide -24 and -6 to simplify this. this is the same answer as 24/6 since two negatives that are divided by each other equal a positive number, and 24 divided by 6 is 4, so the answer is 4.

5 0

3 years ago

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

(9x2 - 8x + 4) - (4x2 + 3x - 9) A. 13x2 - 11x + 5 B. 13x2 - 11x + 13 C. 5x2 - 5x + 13 D. 5x2 - 11x + 13

Troyanec [42]

The answer is d) 5x2-11x+13

4 0

3 years ago

Johnathan plans on using 2" x 12" boards that cost $2.50 per foot. Assuming he can purchase the exact amount he needs, how much

Kitty [74]

Answer:

$35

Step-by-step explanation:

7 0

3 years ago

Equation for ( 6, 13) ( 7900, 5 )

tia_tia [17]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y=mx+b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}$

Thus, the equation of the line is of the form:

$y = - \frac {4} {3947} x + b$

We substitute one of the points and find b:

$13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}$

Finally, the equation is:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

Answer:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

5 0

3 years ago