Answer:
Step-by-step explanation: A on edge
Let, x = The cost of an apple
y = The cost of a peach
6x + 9y = 7.86 - - - - - - - - 1
4x + 5y = 4.82 - - - - - - - - 2
From equation 2,
4x =4.82 - 5y
x= (4.82 - 5y) ÷ 4 - - - - - - - 3
Substitute 3 into 1,
6x + 9y = 7.86
6 [(4.82 - 5y) ÷ 4] + 9y = 7.86
(28.92 - 30 y) ÷4 + 9y = 7.86
28.92 - 30y + 36y = 31.44
6y=2.52
y= 0.42
Substitute y = 0.42 into 2
4x + 5y= 4.82
4x + 5 (0.42) = 4.82
4x + 2.1 = 4.82
4x= 2.72
x= 0.68.
Thus, an apple costs $ 0.68 and a peach costs $ 0.42.
The correct answer is the last option.
Since the coach is studying the two different groups, the results will be more meaningful if the groups are compared to one another as opposed to the players. It would also make more sense to study the mean (or average) of the points scored, rather than the standard deviation (which is essentially a measure of how spread out a set of numbers are).
Answer:
B
Step-by-step explanation:
Answer:
The correct option is;
c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.
Step-by-step explanation:
Here we have the values
μ = 275 mL
275.4
276.8
273.9
275
275.8
275.9
276.1
Sum = 1928.9
Mean (Average), = 275.5571429
Standard deviation, s = 0.921696159
We put the null hypothesis as H₀: μ₁ = μ₂
Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂
The t-test formula is as follows;

Plugging in the values, we have,
Test statistic = 1.599292
at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912
Our p-value from the the test statistic = 0.1608723≈ 0.1609
Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.