Answer:
Step-by-step explanation:
Hello!
To determine whether boys excel in math classes compared to girls, two random samples were collected:
Sample 1
X₁: score achieved by a boy in calculus
n₁= 15
X[bar]₁= 82.3%
S₁= 5.6%
Sample 2
X₂: score obtained by a girl in calculus
n₂= 12
X[bar]₂= 81.2%
S₂= 6.7%
To estimate a confidence interval for the difference between the average percentages of boys and girls in calculus, it's essential that both variables come from normally distributed populations.
For utilizing a pooled variance t-test, it is also required that the population variances, though unknown, are assumed to be equal.
The confidence interval can then be calculated with:
[(X[bar]_1 - X[bar]₂) ± 
* 
]
[(82.3 - 81.2) ± 1.708 * (6.11 * 
]
[-2.94; 5.14]
Using a 90% confidence level, the interval [-2.94; 5.14] is expected to encompass the true difference between the average percentages achieved by boys and girls in calculus.
I hope this is of assistance!
Answer:
The lines are parallel
Step-by-step explanation:
it's established that
To find the slope between two points, we use the formula
Keep in mind
If lines are parallel, their slopes must be identical
When lines are perpendicular, their slopes are negative reciprocals (the product equals -1)
step 1
Determine the slope of JK
we have
insert the values
step 2
Compute the slope of LM
we have
insert the values
step 3
Analyze the slopes
thus
hence
The lines are parallel
A) The product of x and y equals 196. Thus, A) x is equal to 196 divided by y.
B) The sum of x and y is 35. By substituting A) into B), we have
B) 196 divided by y plus y equals 35. Multiplying both sides by y yields
B) 196 plus y squared equals 35y. This results in B) y squared minus 35y plus 196 equals 0.
Therefore, the solutions are X1 = 28 and X2 = 7.
The correct answer is A. According to the question's details, we're provided with key statistics on zucchini weights, which suggest that the average turns out to be typically 0.8 pounds, while the standard deviation is noted as 0.25 pounds. The probability of a randomly selected zucchini weighing between 0.55 pounds and 1.3 pounds can be mathematically expressed. Observing the provided normal distribution options, option A aligns with our specified weight range.
