Question 13 of 25

A ball is chosen at random from a bag containing 150 balls that are either red or blue and either dull or shiny. There are 36 re

tino4ka555 [31]

Answer:

The probability of the chosen ball being shiny conditional on it being red is; 0.375

Step-by-step explanation:

Let A be the event that a red ball has been chosen

Let B be the event that a shiny ball has been chosen

Let S be the total outcomes = 150 balls

Thus;

P(A ∩ B ) = 36/150

A ∩ B' = 150 - 36 - 54

A ∩ B' = 60

Thus; P(A ∩ B') = 60/150

P(A') = 54/150

P(A) = (150 - 54)/150 = 96/150

Thus, probability of the chosen ball being shiny conditional on it being red is;

P(B | A) = P(B ∩ A)/P(A)

Thus; P(B | A) = (36/150)/(96/150)

P(B | A) = 0.375

8 0

2 years ago

Bomdas can someone help me with this 32-7×4+42÷6×3

monitta

25, 7*4 = 28 32-28=4 and 42/6=7*3 = 21 + 4 = 25

6 0

3 years ago

Read 2 more answers

I really need help! ANYONEEEEE

antiseptic1488 [7]

Answer:

I think Its 47? if Im wrong Im very sorry.

Step-by-step explanation:

I added all of the numbers, and the sides without numbers, its simple just add all of the numbers together, I think

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

A Machine can produce six yards of yarn in two minutes. How many yards of yarn can it produce in one hour

kaheart [24]

Answer:180

Step-by-step explanation:

8 0

4 years ago