Bot Math Questions
1. What is the square root of 144?
2. What is the value of π?
3. What is the sum of 7 and 9?
4. What is 3 multiplied by 5?
5. What is the quotient of 24 and 3?
6. What is the difference between 15 and 8?
7. What is 2 to the power of 4?
8. What is 10 percent of 200?
9. What is the circumference of a circle with a diameter of 6 meters?
10. If a ball is thrown 20 meters in 4 seconds, what is its speed?
If the table below represents a probability model based on observed frequencies, what is the value of x?
Y C A R E S
P(Y) 0.25 0.18 0.30 0.17 x
To determine the value of x, we must remember that the sum of all the probabilities in a probability model must equal 1. Therefore, we can set up the equation:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Simplifying this equation, we get:
0.90 + x = 1
Subtracting 0.90 from both sides, we get:
x = 0.10
Therefore, the value of x in this probability model is 0.10.
Thanks!
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Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
To find the estimated difference in the expected number of white and red balls in Bowl B, we need to find the expected number of white and red balls selected from Bowl B and subtract them.
First, we need to find the proportion of white and red balls in Bowl B. From the table, we know that Bowl B contains 60 white balls and 40 red balls out of a total of 100 balls. Therefore, the proportion of white balls in Bowl B is:
60/100 = 0.6
And the proportion of red balls in Bowl B is:
40/100 = 0.4
Next, we need to use the proportions we found to estimate the expected number of white and red balls selected from Bowl B. Since there are 500 balls in Bowl B, we can estimate that Clark will select 500 balls from Bowl B in total.
The expected number of white balls selected from Bowl B is:
0.6 * 500 = 300
And the expected number of red balls selected from Bowl B is:
0.4 * 500 = 200
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is:
300 - 200 = 100
So the estimated difference in expected number of white and red balls in Bowl B is 100.
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.
The total number of flowers in the vase is:
5 + 6 + 7 = 18
The probability of selecting the first carnation is:
6/18
After the first carnation is selected, there are 5 carnations and 17 flowers left. The probability of selecting a second carnation is:
5/17
Since the flowers are selected without replacement, we need to multiply the probabilities of each event to get the overall probability of selecting two carnations:
(6/18) * (5/17) = 30/306
Simplifying this fraction, we get:
5/51
Therefore, the probability of selecting two carnations is 5/51.