Answer:
160/1001, 175/1001
Step-by-step explanation:
i) We calculate:
₈C₁ methods to select 1 new camera from a selection of 8
₆C₃ methods to select 3 refurbished cameras from a selection of 8
₁₄C₄ methods to select 4 cameras from the total of 14 cameras
The probability formula is:
P = ₈C₁ ₆C₃ / ₁₄C₄
P = 8×20 / 1001
P = 160 / 1001
P ≈ 0.160
ii) For at most one new camera, it means we want either one new camera or none at all. We've calculated the probability of selecting one new camera already. The probability of not selecting any new camera is equivalent to selecting 4 refurbished cameras:
P = ₆C₄ / ₁₄C₄
P = 15 / 1001
Therefore, the combined probability is:
P = 160/1001 + 15/1001
P = 175/1001
P ≈ 0.175
To derive the function that characterizes the bee population:
1) Initially, there are 9,000 bees in the first year.
2) In the second year, a reduction of 5% occurs => 9,000 - 0.05 * 9,000 = 9,000 * (1 - 0.05) = 9,000 * 0.95
3) Each subsequent year sees a 5% decline => 9,000 * (0.95)^(number of years)
4) Let x represent years and f(x) signify the bee count, then: f(x) = 9,000 (0.95)^x.
Evaluation of the claims:
<span>1) The function f(x) = 9,000(1.05)x applies to the scenario.
FALSE: WE ESTABLISHED IT AS f(x) = 9,000 (0.95)^x
2) The function f(x) = 9,000(0.95)x applies to the scenario.
TRUE: THIS IS THE RESULT OF OUR PRIOR ANALYSIS.
3) After 2 years, the farmer projects approximately 8,120 bees will be left.
Calculating:
f(2) = 9,000 * (0.95)^2 = 9,000 * 0.9025 = 8,122
Thus, this statement is TRUE
4) After 4 years, the farmer can predict there will be roughly 1,800 bees left.
f(4) = 9,000 * (0.95)^4 = 9,000 * 0.81450625 = 7,330
This statement is therefore FALSE
5) The domain values contextual to this situation are restricted to whole numbers.
FALSE: DOMAIN VALUES INCLUDE ALL NON-NEGATIVE REAL NUMBERS. FOR INSTANCE, THE FUNCTION IS VALID AT X = 5.5
6) The range values pertinent to this situation are restricted to whole numbers.
TRUE: FRACTIONS OF BEES CANNOT EXIST.
</span>
Answer:
1. 3.767
2. 0.145
Step-by-step explanation:
Define X as the exam scores and Y as the number of drinks.
X Y X-Xbar Y-Ybar (X-Xbar)(Y-Ybar) (X-Xbar)² (Y-Ybar)²
75 5 -2.3 2.3 -5.29 5.29 5.29
92 3 14.7 0.3 4.41 216.09 0.09
84 2 6.7 -0.7 -4.69 44.89 0.49
64 4 -13.3 1.3 -17.29 176.89 1.69
64 2 -13.3 -0.7 9.31 176.89 0.49
86 7 8.7 4.3 37.41 75.69 18.49
81 3 3.7 0.3 1.11 13.69 0.09
61 0 -16.3 -2.7 44.01 265.69 7.29
73 1 -4.3 -1.7 7.31 18.49 2.89
93 0 15.7 -2.7 -42.39 246.49 7.29
sumx=773, sumy=27, sum(x-xbar)(y-ybar)= 33.9, sum(X-Xbar)²= 1240.1,sum(Y-Ybar)²= 44.1
Xbar=sumx/n=773/10=77.3
Ybar=sumy/n=27/10=2.7
1.
Cov(x,y)=33.9/9
Cov(x,y)=3.76667
Thus, the sample covariance of exam scores and energy drink consumption is 3.767
2.
Cor(x,y)=r=33.9/233.85553
Cor(x,y)=r=0.14496
The sample correlation coefficient for the relationship between exam scores and energy drink consumption is 0.145.
