The weight (in pounds) for a population of school-aged children is normally distributed with a mean equal to 133 ± 25 pounds (μ ± σ). suppose we select a sample of 100 children (n = 100) to test whether children in this population are gaining weight at a 0.05 level of significance. part (a) what are the null and alternative hypotheses? h0: μ ≤ 133 h1: μ > 133 h0: μ = 133 h1: μ < 133 h0: μ = 133 h1: μ ≠ 133 h0: μ ≤ 133 h1: μ = 133

Null hypothesis:    

Alternative hypothesis :  

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

Data given and notation  

represent the mean breaking strength value for the sample  

represent the population standard deviation  

sample size  

represent the value that we want to test  

represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

Is a one right tailed test.  

What are H0 and Ha for this study?  

We want to test if the children in this population are gaining weight, so we want to test if the mean increase from the reference value or no.

Null hypothesis:    

Alternative hypothesis :  

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

8+10

step-by-step explanation:

you multiply 4 by 2 and 5 by 2