Answer:
First Choice: As the number of hours spent on homework increases, the tests scores increase.
Step-by-step explanation:
The definition of a positive correlation is a relationship between two given variables, in which both variables are moving in the same direction. This can mean when one variable increases and the other variable increases, too, or one variable decreases and the other decreases as well.
The first choice is a positive correlation because both variables are changing (increasing) in the same direction. As you spend more time on homework, you're likely to get a higher test score.
The second choice cannot be a positive correlation because only one variable is having some kind of change (increasing). The doctor visits amount remains the same, so we can call this a zero-correlation relationship because the number of apples eaten yearly doesn't affect the amount of doctor visits. An apple a day keeps the doctor a way is just a proverb, not to be taken literally.
The third choice cannot be a positive correlation because the two variables are going different directions. Even though the number of times going to bed early is increasing, the number of times waking up late decreases, which is not moving in the same direction as the other variable.
The fourth choice cannot be a positive correlation because, similarly to the third choice, the two variables are going different directions. One variable is increasing, which is the amount of practice time. Meanwhile, the other variable is decreasing (going in the opposite direction), which is the number of games lost in a season.
84 x .70 = $58.80
Or if you want to do it the long way you can calculate the actual discount (your savings from this purchase first then subtract it from the original price of $84. It`s just easier to figure out what is left (which is 70% of $84).
Answer:
(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0
(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0
Step-by-step explanation:
(a) when i = 2, the expected number of played games will be:
E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] = 4p²-4p+2-6p²+6p = -2p²+2p+2.
If p = 1/2, then:
d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.
(b) when i = 3;
E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]
Simplification and rearrangement lead to:
E(X) = 6p⁴-12p³+3p²+3p+3
if p = 1/2, then:
d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10
Therefore, E(X) is maximized.
Answer:
a) Bar chart
b) Histogram
c) Bar chart
d) Histogram
Step-by-step explanation:
a) Trash pick-up DAY for each HOUSEHOLD in Ames - This is categorical data because, we are talking about days of the week. For instance, Household 1 might have Sunday as Trash pick-up day and that could be accumulated into frequency. Hence, BAR chart is the most appropriate.
b) Patient WAIT-TIME at ISU. This is continuous (quantitative) data. And the most appropriate is HISTOGRAM.
c) Number of trips taken during a GIVEN SCHOOL YEAR by EACH ISU STUDENT. Let say we have 5 ISU STUDENTS. Student 1 had 5 trips, student 2 had 10 trips, etc.
We want to see which student has the most and least trip in that particular school year. Although is count data but the most appropriate graphical display is BAR chart.
d) TAX BRACKET of ALL Iowa RESIDENTS. This is a continuous (quantitative) data. The most appropriate graphical display is HISTOGRAM.
