Answer:
Step-by-step explanation:
Hello!
To test the claim that eating a healthy breakfast improves the performance of students on their test a math teacher randomly asked 46 students what did they have for breakfast before they took the final exam and classified them as:
<u>Group 1</u>: Ate healthy breakfast
X₁: Number of students that ate a healthy breakfast before the exam and earned 80% or higher.
n₁= 26
<u>Group 2: </u>Did not eat healthy breakfast
X₂: Number of students that did not eat a healthy breakfast before the exam and earned 80% or higher.
n₂= 20
After the test she counted the number of students that got 80% or more in the test for each group obtaining the following sample proportions:
p'₁= 0.50
p'₂= 0.40
The parameters of study are the population proportions, if the claim is true then p₁ > p₂
And you can determine the hypotheses as
H₀: p₁ ≤ p₂
H₁: p₁ > p₂
α: 0.05
≈N(0;1)
pooled sample proportion: 
![Z_{H_0}= \frac{(0.5-0.4)-0}{\sqrt{0.46(1-0.46)[\frac{1}{26} +\frac{1}{20}] } } }= 0.67](https://tex.z-dn.net/?f=Z_%7BH_0%7D%3D%20%5Cfrac%7B%280.5-0.4%29-0%7D%7B%5Csqrt%7B0.46%281-0.46%29%5B%5Cfrac%7B1%7D%7B26%7D%20%2B%5Cfrac%7B1%7D%7B20%7D%5D%20%7D%20%7D%20%7D%3D%200.67)
p-value: 0.2514
The decision rule is:
If p-value ≤ α, reject the null hypothesis.
If p-value > α, do not reject the null hypothesis.
The p-value: 0.2514 is greater than the significance level 0.05, the test is not significant.
At a 5% significance level you can conclude that the population proportion of math students that obtained at least 80% in the test and had a healthy breakfast is equal or less than the population proportion of math students that obtained at least 80% in the test and didn't have a healthy breakfast.
So having a healthy breakfast doesn't seem to improve the grades of students.
I hope this helps!
Pair 1: slope = (9 - 5)/(8+4) = 1/3
midpoint = ((-4+8)/2, (5+9)/2) = (2, 7)
perpendicular bisector passes through point (2, 7) with slope = -1/(1/3) = -3 giving the equation (y - 7)/(x - 2) = -3 or y - 7 = -3(x - 2) or y = -3x + 13 and y-intercept at y = 13.
Pair 2: slope = (6 - 4)/(-8-2) = -1/5
midpoint = ((2-8)/2, (4+6)/2) = (-3, 5)
perpendicular
bisector passes through point (-3, 5) with slope = -1/(-1/5) = 5 giving
the equation (y - 5)/(x + 3) = 5 or y - 5 = 5(x + 3) or y = 5x + 20
and y-intercept at y = 20.
Pair 3: slope = (2 - 4)/(7 - 5) = -1
midpoint = ((5+7)/2, (4+2)/2) = (6, 3)
perpendicular
bisector passes through point (6, 3) with slope = -1/(-1) = 1 giving
the equation (y - 3)/(x - 6) = 1 or y - 3 = (x - 6) or y = x - 3
and y-intercept at y = -3.
Pair 4: slope = (3 - 9)/(-4 - 2) = 1
midpoint = ((2-4)/2, (9+3)/2) = (-1, 6)
perpendicular
bisector passes through point (-1, 6) with slope = -1(1) = -1 giving
the equation (y - 6)/(x + 1) = -1 or y - 6 = -1(x + 1) or y = -x + 5
and y-intercept at y = 5.
Pair 5: slope = (-12 + 2)/(9 - 3) = -5/3
midpoint = ((3+9)/2, (-2-12)/2) = (6, -7)
perpendicular
bisector passes through point (6, -7) with slope = -1(-5/3) = 3/5 giving
the equation (y + 7)/(x - 6) = 3/5 or 5(y + 7) = 3(x - 6) or 5y = 3x - 53
and y-intercept at y = -10.6.
Pair 6: slope = (12 - 10)/(8 - 4) = 1/2
midpoint = ((4+8)/2, (10+12)/2) = (6, 11)
perpendicular
bisector passes through point (6, 11) with slope = -1(1/2) = -2 giving
the equation (y - 11)/(x - 6) = -2 or y - 11 = -2(x - 6) or y = -2x + 23
and y-intercept at y = 23.
Arrangement in order of y-intercepts from smallest to largest
a(3, -2) and b(9, -12)
a(5, 4) and b(7, 2)
a(2, 9) and b(-4, 3)
a(-4, 5) and b(8, 9)
a(2, 4) and b(-8, 6)
a(4, 10) and b(8, 12)
The simplest way to do this is to set up equivalent fractions. So, you'd do:

=

and then solve for x.
x=52
ANSWER

EXPLANATION
The given function is

To find f(-9), we substitute x=-9 into the function to obtain:

Simplify:

Simplify further;

Answer:
About 0.72
Step-by-step explanation:
Given the data:
With device : 22.6, 23.4, 28.4, 29, 29.3, 20.0
Without device :26.9, 24.4, 20.8, 20.8, 20.2, 26.0, 28.1, 25.6
Using calculator:
With device data:
Sample size (n1) = 6
Degree of freedom (df1) = 6 - 1 = 5
Mean(m1) = 25.45
Variance (s²1) = 15.631
Without device data:
Sample size (n2) = 8
Degree of freedom (df2) = n - 1 = 7
Mean (m2) = 24.1
Variance (s²2) = 9.54
s²t = ((df1/(df1 + df2)) * s²1) + ((df2/(df1 + df2)) * s²2)
s²t = ((5/(5+7)*15.63) + ((7/(5+7))*9.54)
= ((5/12) * 15.63) + ((7/12) * 9.54) = 12.0775
s²m1 = s²t/n1 = 12.0775/6 = 2.0123
s²m2 = s²t/n2 = 12.0775/8 = 1.511
T - statistic :
Tstat = (m1 - m2)/√(s²m1 + s²m2)
Tstat = (25.45 - 24.10) / √(2.0123 + 1.511)
Tstat= 1.350/√3.5233
Tstat = 1.350 / 1.8770455
Tstat = 0.7192153 = 0.72