Step-by-step explanation:
When a negative number is placed within a modulus function, the result will be positive. For instance, |-3| equals 3, |-6| equals 6, and |5| equals 5, etc.
A modulus function, expressed as |x|, is always positive unless x is zero, in which case it equals zero.
Consequently, |x| cannot be less than -4 because |x| is always non-negative. Thus, the statement is inaccurate.
Answer:
Comprehensive explanation:
Let's denote
x -----> the number of days
and
y ----> the remaining minutes for Yuson
We know that
The linear equation in slope-intercept form can be represented as
here
m indicates the slope
b represents the y-coordinate of the y-intercept (starting value)
In this scenario, we have
The slope is defined as
----> negative, as it indicates a decreasing function
----> initial value
substituting the values
1.4×5=7
0.8×10=8
1.4×10=14
1×15=15
15+14+8+7=44
44÷4=11
LQ of 44=11
LQ=10 minutes
11×3=33
UQ= 29 minutes
The Range is 19 minutes
Detailed breakdown:
Commence with the individual boxes. For determining the number of students in each category, calculate Frequency density × The difference in the category. (if it's 5-15, the difference is 10)
This results in the counts of students in each range.
Next, determine the LQ of 44, which is 11.
Then locate the 11th student's score; in this instance, it resides in the 5-15 range. 7 students have already surpassed it, with 8 in the 5-15 range. Hence, the 11th lies within the bounds of 5-15, making the middle 10.
Repeat this process for the UQ.
The interquartile range is calculated as UQ-LQ, yielding 29-10=19 minutes.
I hope this helps, though I'm not entirely sure if my explanation is coherent and I'm unclear on the terminology I've used for these categories.
Answer:
a) Ann has a 1/3 chance of winning in the first round
b) The chance of Ann winning for the first time in the fourth round is 8/81
c) The probability that Ann's first win occurs after the fourth round is 16/81
Step-by-step explanation:
a) Each strategy is played with a probability of 1/3. Given any strategy, there’s a 1/3 chance that Bill will choose the strategy that allows Ann to win. Consequently, the probability of Ann securing a victory in the first round (or any round) is
1/3 * 1/3 + 1/3 * 1/3 + 1/3 * 1/3 = 1/9 + 1/9 + 1/9 = 1/3.
Thus, the likelihood of Ann winning the initial round is 1/3.
b) The chances of Ann winning a round stand at 1/3; therefore, her chances of not winning are 2/3. This must happen three times before her first victory. Thus, the probability that Ann's first win occurs in the fourth round is
(2/3)³ * 1/3 = 8/81.
c) The first victory happens after the fourth round if she remains unsuccessful in the first four rounds, translating to a possibility of (2/3)⁴ = 16/81.
The P-value to evaluate the claim that the mean length of pencils produced in this factory equals 18.0 cm is 0.00736. Step-by-step explanation: In this case, a quality control specialist extracted a random sample of 45 pencils from the assembly line, which exhibited a mean length of 17.9 cm. With a known population standard deviation of 0.25 cm, we denote by the population mean length for pencils produced in the factory. Thus, Null Hypothesis: = 18.0 cm (indicating that the population mean length equals 18.0 cm). Alternate Hypothesis: ≠ 18.0 cm (suggesting different from 18.0 cm). We apply the one-sample z-test since the population standard deviation is known. The test statistic yields: T.S. ~ N(0,1), with the sample mean length 17.9 cm and population standard deviation 0.25 cm for a sample size of 45. Hence, the calculated test statistic is -2.68. The corresponding P-value is derived from P(Z < -2.68) = 1 - P(Z > 2.68), equating to 1- 0.99632 = 0.00368. For a two-tailed test, the resulting P-value computes to 2 * 0.00368 = 0.00736.
