If three out of every fourteen trick-or-treaters that came to your house last Halloween were dressed as pirates, what proportion of trick-or-treaters were not dressed as pirates?

Answers

Answer 1
Answer:

Answer:

Explanation:

Given that three out of every fourteen trick-or-treaters were dressed as pirates

The proportion of the tick-or-treaters that were not dressed as pirates is the subtraction of the proportion of the people d

Dressed as pirates = 3/14

Not dressed as pirates = 14/14 - 3/14

= 11/14


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PLEASE ANSWER Add. 51.342 + 36.530

Answers

Use the calculator :v nvm the answer is 87.872
87.872 is the answer

Determine the measure of < 7
o 53
o 133
o 127
o 47

Answers

Answer:
• 127

Explanation:
How to find 127,
The line is 180°, so you take 53 minus 180.
180-53=127

Hope this helps! Good luck :)

Suppose that B1 and B2 are mutually exclusive and complementary events, such that P(B1 ) = .6 and P(B2) = .4. Consider another event A such that P(A | B1) = .2 and P(A | B2) = .5. Find P(A).

Answers

Answer:

So, we get that is P(A)=0.32.

Step-by-step explanation:

We know that:

P(B_1)=0.6\n\nP(B_2)=0.4\n\nP(A|B_1)=0.2\n\nP(A|B_2)=0.5\n

We have the formula for probability:

P(A|B)=(P(A\cap B))/(P(B))\n\n\implies P(A\cap B)=P(A|B)\cdot P(B)

So, we calculate:

P(A\cap B_1)=P(A|B_1)\cdot P(B_1)\n\nP(A\cap B_1)=0.2\cdot 0.6=0.12\n\n\nP(A\cap B_2)=P(A|B_2)\cdot P(B_2)\n\nP(A\cap B_2)=0.5\cdot 0.4=0.2\n

We calculate:

P(A)=P((A\cap B_1)\cup(A\cap B_2))\n\nP(A)=P(A\cap B_1)+P(A\cap B_2)\n\nP(A)=0.12+0.2\n\nP(A)=0.32

So, we get that is P(A)=0.32.

Final answer:

To find P(A), use the law of total probability given that B1 and B2 are mutually exclusive and complementary events. Substituting the provided values, P(A) = 0.32.

Explanation:

The question is asking us to calculate P(A), given the values for P(A | B1) and P(A | B2), and the knowledge that B1 and B2 are mutually exclusive and complementary events. In probability, if events B1 and B2 are mutually exclusive and complementary, this means that one and only one of them can occur, and their occurrence covers all possible outcomes. We can use the law of total probability to find the overall P(A). The law of total probability states that P(A) = P(A | B1) * P(B1) + P(A | B2) * P(B2). Plugging the provided values into this formula, we get P(A) = .2 * .6 + .5 * .4 = .12 + .2 = .32. Therefore, P(A) is .32.

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Read the statement shown below.If Amelia finishes her homework, then she will go to the park.

Which of these is logically equivalent to the given statement? (1 point)


1. If Amelia did not go to the park, then she did not finish her homework.
2. If Amelia did not finish her homework, then she will go to the park.
3. If Amelia goes to the park, then she did not finish her homework.
4. If Amelia finishes her homework, then she cannot go to the park.

Answers

Hello there.

In this problem, we can use our intuition of logic, but I will show a proof of the result in a truth table later. Then, let's get started!

Given:

→ Amelia finishes the homework (sentence H, can be True or False)

→ Amelia goes to the park (P, true or false)

Then, we have: If H, then P. Logically:

H ⇒ P

Then we can think: everytime she does the homework, she goes to the park. Therefore, if she did not go to the park, she will not have finished the homework (It is an equivalent sentence).

Alternative 1.

==========

Now, let's prove that (H ⇒P) is equivalent to (¬P ⇒ ¬H), via the truth table:

H P ¬H ¬P (H ⇒ P) (¬P ⇒ ¬H)

T T F F T T

T F F T F F

F T T F T T

F F T T T T

As we can see, the results are identical, therefore, the sentences are indeed equivalent.

I hope it hepls :)

Number 1 is the same thing but told differently.

(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $42,500 with a standard deviation of $6,800. Find an 90% confidence interval for ????. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: 41101.87 Upper-bound: 43898.13

Answers

Answer:

41101.750 to 43898.250

Step-by-step explanation:

Using this formula X ± Z (s/√n)

Where

X = 42500 --------------------------Mean

S = 6800----------------------------- Standard Deviation

n = 64 ----------------------------------Number of observation

Z = 1.645 ------------------------------The chosen Z-value from the confidence table below

Confidence Interval Z

80%. 1.282

85% 1.440

90%. 1.645

95%. 1.960

99%. 2.576

99.5%. 2.807

99.9%. 3.291

Substituting these values in the formula

Confidence Interval (CI) = 42500 ± 1.645(6800/√64)

CI = 42500 ± 1.645(6800/8)

CI = 42500 ± 1.645(850)

CI = 42500 ± 1398.25

CI = 42500+1398.25 ~. 42500-1398.25

CI = 43898.25 ~ 41101.75

In other words the confidence interval is from 41101.750 to 43898.250

Final answer:

To find a 90% confidence interval for the mean starting salary, use the formula CI = sample mean ± (Z * sample standard deviation / √n).

Explanation:

To find a 90% confidence interval for the mean starting salary, we will use the formula:

CI = sample mean ± (Z * sample standard deviation / √n)

Given that the sample mean is $42,500, the sample standard deviation is $6,800, and the number of college graduates is 64, we can substitute these values into the formula to calculate the confidence interval. The lower-bound is $41,101.87 and the upper-bound is $43,898.13.

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HELP ME PLEASE I DO NOT UNDERSTAND PLEASE THANK YOU HAVE A GOOD DAYa.) Draw a rectangular prism with these dimensions: 3 units x 9 units x 27 units.






b.) Draw a net of the rectangular prism and label each face with its surface area.












c.) Use the net of the rectangular prism to find the surface area of the solid.

Answers

You can draw a rectangular prison not to scale and just label these on the prism length= 3 width= 9 and height =27
Then you would draw a net the rectangular prism like if you were to unfold it then use the calculation from a and solve for the surface area
Then you add the surface areas together I believe sorry if I get this wrong and it should be the surface area of everything