Answer:
The required division problem he must solve is:
Step-by-step explanation:
Consider the provided information.
Martin chose two of the cards below. When he found the quotient of the numbers, his answer was -16/9.
As we know that the quotient of the number is a negative number.
Therefore, the sign of both numbers must be different,
Thus we can concluded he must select 
as one of the card, so that product is a negative number.
Let the selected card be x.
Hence, the two cards should be and 
The required division problem he must solve is:
Answer:
jump discontinuity at x = 0; point discontinuities at x = –2 and x = 8
Step-by-step explanation:
From the graph we can see that there is a whole in the graph at x=-2.
This is referred to as a point discontinuity.
Similarly, there is point discontinuity at x=8.
We can see that both one sided limits at these points are equal but the function is not defined at these points.
At x=0, there is a jump discontinuity. Both one-sided limits exist but are not equal.
Answer:
<h2>One</h2>
Step-by-step explanation:
Given the value 78247 a s a stable number because at least one of its digits has the same value as its position in the number. The 4th number in the value is 4, this makes the number a stable number.
The following are the 3-digits stable numbers that appears in 78247
The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).
Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.
Answer:
Part A
Please see attached the required stem and leaf plot
For the stem and leaf plot, the nonsplit system is used because of clarity for analysis
Part B:
From the shape of the stem and leaf plot we have that there is an average increase of pulse rate of 20 pulses in all the 19 students after the exercise
The shape of the plot is relatively the same for the before and after exercise save for the decrease in the third to the last row by one and the increase in the second to the last roe by one student
The spread remained relatively constant in both cases with the most being in the 60s range having 7 students in the before exercise and the 80s range having 8 students in the after exercise leaf plot.
Step-by-step explanation:
The given data are;
67
87
67 88
67 89
68 89
71 91
72 93
72 93
75 95
77 96
77 97
79 98
81 98
85 101
87 105
87 105
91 119
97 125
103 125
121 147
Answer:
The standard deviation of the number of rushing yards for the running backs that season is 350.
Step-by-step explanation:
Consider the provided information.
The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards.
Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.
Use the formula: 
Here z is 2.42 and μ is 790, substitute the respective values as shown.
Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.
