Answer:
The confidence interval is from 69.82 o 82.18
Step-by-step explanation:
Using this formula X ± Z (s/√n)
Where
X = 76 --------------------------Mean
S = Standard Deviation
If Variance = 144
S = √144
S = 12
n = 25 ----------------------------------Number of observation
Z = 2.576 ------------------------------The chosen Z-value from the confidence table below
Confidence Interval || Z
80%. || 1.282
85% || 1.440
90%. || 1.645
95%. || 1.960
99%. || 2.576
99.5%. || 2.807
99.9%. || 3.291
Substituting these values in the formula
Confidence Interval (CI) = 76 ± 2.576 (12/√25)
CI = 76 ± 2.576(12/5)
CI = 76 ± 2.576(2.4)
CI = 76 ± 6.1824
CI = 76 + 6.1824 ~ 76 - 6.1824
CI = 82.1824 ~ 69.8176
CI = 82.18 ~ 69.82
In other words the confidence interval is from 69.8176 to 82.1824
Answer:
I'm a little confused on what is being asked to do.
Step-by-step explanation:
Answer:
6 units
Step-by-step explanation:
Remember the formula for the area of a rectangle: A = lw
What we know:
A=48
w = l-2
Substitute A for 48 and w for l-2 into the equation
A = lw
48 = l(l-2) Use the distributive property. Multiply over the brackets.
48 = l² - 2l
Rearrange the equation to standard form (0 = ax² + bx + c) to use quadratic formula.
0 = l² - 2l - 48
a = 1 ; b = -2 ; c = -48 State the variables for the quadratic formula
Substitute a, b and c to find the length:

Simplify


Split the equation at the ± for adding and subtracting. Then decide which answer is correct, or if both of them are possible answers.


This is "inadmissable", or impossible because the length can't be a negative value.


Length of the rectangle
Use the formula for the area of a rectangle
Substitute the length and area, then isolate "w" for the width
A = lw
48 = (8)w
48/8 = w
w = 6
Therefore the length of the rectangle is 6 units.
Hello there,
Slope:
6 - (-5) / -6 - 0
11/-6
<em>Slope=</em> -11/6
<em>Slope intercept form: </em>y = mx + b
-5 = -11/6(0) + b
-5 = b
Answer: y = -11/6x - 5
Hope I Helped!
-Char
Answer:
the probability is P=0.012 (1.2%)
Step-by-step explanation:
for the random variable X= weight of checked-in luggage, then if X is approximately normal . then the random variable X₂ = weight of N checked-in luggage = ∑ Xi , distributes normally according to the central limit theorem.
Its expected value will be:
μ₂ = ∑ E(Xi) = N*E(Xi) = 121 seats * 68 lbs/seat = 8228 lbs
for N= 121 seats and E(Xi) = 68 lbs/person* 1 person/seat = 68 lbs/seat
the variance will be
σ₂² = ∑ σ² (Xi)= N*σ²(Xi) → σ₂ = σ *√N = 11 lbs/seat *√121 seats = 121 Lbs
then the standard random variable Z
Z= (X₂- μ₂)/σ₂ =
Zlimit= (8500 Lbs - 8228 lbs)/121 Lbs = 2.248
P(Z > 2.248) = 1- P(Z ≤ 2.248) = 1 - 0.988 = 0.012
P(Z > 2.248)= 0.012
then the probability that on a randomly selected full flight, the checked-in luggage capacity will be exceeded is P(Z > 2.248)= 0.012 (1.2%)