The table shows the results of drawing letter tiles from the bag. What is the probability that the next tile drawn will have the
letter B on it?
A. 1/10
B. 1/5
C. 2/5
D. 3/5
A. 1/10
B. 1/5
C. 2/5
D. 3/5
1 answer:
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Answer:
Senn needs to give Ms. Lady 4.6 minutes head start
Senn needs to give Ms. Beetle 18.4 minutes head start
The equation for each bug are;
For Senn, the equation for the experiment is 5 × t₁ = 92 feet
For Ms. Lady, the equation for the experiment is 4 × (t₁ + 4.6) = 92 feet
For Ms. Beetle , the equation for the experiment is 4 × (t₁ + 18.4) = 92 feet
Please find attached the required graph and table of values
Step-by-step explanation:
The given parameters are;
The average crawling time of Ms. Lady = 4 feet per minute
The average crawling time of Ms. Beetle = 2.5 feet per minute
The average crawling time of Senn = 5 feet per minute
The distance to the rose bush = 92 feet
Therefore, we have;
The time, t, duration for Senn to arrive at the Rose bush is given by the following relation,
Time, t = Distance, d/(Speed, s)
Given that the speed of the bugs is equal to their average crawling time, we have
For Senn
Time, t = 92/(5 ft/Min) = 18.4 minutes
For, Ms. Lady
Time, t = 92/(4 ft/Min) = 23 minutes
For, Ms. Beetle
t = 92/(2.5 ft/Min) = 36.8 minutes
Therefore;
Senn needs to give Ms. Lady 23 - 18.4 = 4.6 minutes head start
Similarly, Senn needs to give Ms. Beetle 36.8 - 18.4 = 18.4 minutes head start
The equation for each bug are therefore;
For Senn, the equation for the experiment is given as follows ;
5 × t₁ = 92 feet
For Ms. Lady, the equation for the experiment is given as follows;
4 × (t₁ + 4.6) = 92 feet
For Ms. Beetle , the equation for the experiment is given as follows;
4 × (t₁ + 18.4) = 92 feet.
