Can someone help i don't understand this
1 answer:
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QUESTION 4
The given system of linear equations are,
and
We put equation (1) into equation (2) to obtain,
We expand to get,
Group like terms to get,
Simplify to get,
Divide both sides by -13 to get,
Put the value of x into equation (1) to get,
The solution is
QUESTION 5
The given system of linear equations are
and
We make y the subject in equation (1) to get,
Put equation (3) into equation (2) to get,
We expand to get,
We group like terms to get,
This implies that,
We divide both sides by -9 to get,
Put the value of x in to equation 3 to get,
Therefore the solution is
QUESTION 6
The given equations are,
and
Make x the subject in equation (2) to get,
Put equation (3) into equation (1) to get,
We expand to get,
We group like terms to obtain,
.Simplify to get,
Divide through by -16 to get,
Put the value of y into equation (3) to get,
Therefore
The given system of linear equations are,
and
We put equation (1) into equation (2) to obtain,
We expand to get,
Group like terms to get,
Simplify to get,
Divide both sides by -13 to get,
Put the value of x into equation (1) to get,
The solution is
QUESTION 5
The given system of linear equations are
and
We make y the subject in equation (1) to get,
Put equation (3) into equation (2) to get,
We expand to get,
We group like terms to get,
This implies that,
We divide both sides by -9 to get,
Put the value of x in to equation 3 to get,
Therefore the solution is
QUESTION 6
The given equations are,
and
Make x the subject in equation (2) to get,
Put equation (3) into equation (1) to get,
We expand to get,
We group like terms to obtain,
.Simplify to get,
Divide through by -16 to get,
Put the value of y into equation (3) to get,
Therefore
Answer:
z = -1.645
Step-by-step explanation:
b) n =25, M = 4.01 , \mu = 4.22 , \sigma = 0.60 , \alpha= 0.05
The hypothesis are given by,
H0 : \mu\geq 4.22 v/s H1 : \mu < 4.22
The test statistic is given by,
Check attachment for the formula that should br here.
z = \frac{4.01- 4.22 }{0.60 /\sqrt{25}}
= -1.75
The critical value of z = -1.645
The calculated value z > The critical value of z
Hence we reject null hypothesis.
The healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population.
