Hace 5 años la edad de un joven era 1/3 de la edad del papá y dentro

g Recall the method used to obtain a confidence interval for the difference between two population means for matched samples. (a

Dovator [93]

Answer:

Step-by-step explanation:

Population 1: 11, 7, 9, 12, 13, 15, 15

Population 2: 6, 6, 4, 5, 8, 13, 12

Mean = sum of terms/number of terms

For population 1,

number of terms = 7

Mean = (11 + 7 + 9 + 12 + 13 + 15 + 15)/7 = 82/7 = 11.7

For population 2,

Number of terms = 7

Mean = (6 + 6 + 4 + 5 + 8 + 13 + 12)/7 = 7.71

The difference value for each element is

(11 + 7 + 9 + 12 + 13 + 15 + 15) - (6 + 6 + 4 + 5 + 8 + 13 + 12) = (5, 1, 5, 7, 5, 2, 3)

The difference between the two means is

Mean of population 1 - mean of population 2

Difference = 11.7 - 7.71 = 3.99

8 0

3 years ago

Peggy earned $20 for each lawn she mowed last summer. This summer, she raised her price to $23 per lawn. What is the percent inc

anyanavicka [17]

Answer:

15 percent

Step-by-step explanation: You will substract the new value from the old value.

23-20

Now you will get the difference between them and divide that by the original amount.

3/20

This will equal 0.15 in which you now multiply by 100

15 percent

5 0

2 years ago

Question 30(Multiple Choice Worth 1 points) (01.01) Given the equation 2x − 5 = 8, which order of operations completely solves f

xeze [42]

Add five to each side, then divide each side by 2. you will get 6.5 in the end

5 0

3 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$