Construct an interval estimate for the given parameter using the given sample statistic and margin of error. For μ1-μ2, using x¯

First note the ratio is
7:2 so divide 7 by 2 to get 7/2 = 3.5. Then add 3.5 to 2, yielding the result 5.5

I will designate the hourly rate for weekdays as x and for weekends as y. The equations are arranged as follows:

13x + 14y = $250.90
15x + 8y = $204.70

This gives us a system of equations which can be solved by multiplying the first equation by 4 and the second by -7. This leads to:

52x + 56y = $1003.60
-105x - 56y = -$1432.90

By summing these two equations, we arrive at:

-53x = -$429.30 --> 53x = $429.30 --> (dividing both sides by 53) x = 8.10. This represents her hourly wage on weekdays.

Substituting our value for x allows us to determine y. I will utilize the first equation, but either could work.

$105.30 + 14y = $250.90. To isolate y, subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40

Thus, we find that her earnings are $8.10 per hour on weekdays and $10.40 per hour on weekends. The difference shows she earns $2.30 more on weekends than on weekdays.

Jacob's lunch account will deplete in 16 days while Samantha's will run out in just 2 days. Step-by-step explanation:

Answer: C. Significant at 0.036

Step-by-step explanation:

Given:

Total samples selected Ns= 500

Airplanes that arrived on time Na = 482.

Airplanes that arrived late Nl = 500 - 482 = 18

Calculating the probability of an airplane arriving late:

P(L) = Nl/Ns

P(L) = 18/500

P(L) = 0.036

An event is deemed significant if its probability is equal to or less than 0.05.

As P(L) < 0.05

P(L) = Significant at 0.036

Answer:

a. Alpha equals 3.014 while beta equals 12.442

b. The likelihood that the data transfer duration surpasses 50ms is 0.238

c. The chance that data transfer time falls between 50 and 75 ms is 0.176

Step-by-step explanation:

a. Given the data, the mean and standard deviation for the random variable X are 37.5 ms and 21.6, respectively.

Thus, E(X)=37.5 and V(X)=(21.6)∧2  

To find alpha, we need to apply the formula:

alpha=E(X)∧2/V(X)

alpha=(37.5)∧2/21.6∧2

alpha=1,406.25 /466.56

​alpha=3.014

To determine beta, the following formula is employed:

β=  V(X) ∧2/E(X)

β=(21.6)  ∧2/37.5

β=466.56 /37.5

β=12.442

b. With E(X)=37.5 and V(X)=(21.6)∧2,  

Hence, P(X>50)=1−P(X≤50)

To find the probability of data transfer time exceeding 50ms, we use the formula:

P(X>50)=1−P(X≤50)

=1−0.762

=0.238

The chance of data transfer time exceeding 50ms is 0.238

c. With E(X)=37.5 and V(X)=(21.6)∧2,  

Thus, P(50<X<75)=P(X<75)−P(X<50)  

To find the probability that data transfer time is between 50 and 75 ms, we apply the formula:

P(50<X<75)=P(X<75)−P(X<50)

=0.938−0.762

=0.176

​

The probability that data transfer time falls between 50 and 75 ms is 0.176