As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

Step-by-step explanation:

Here n= 20

Sample mean GPA = x`= 2.84

Standard mean GPA = u= 2.55

Standard deviation = s= 0.45.

Level of Significance.= ∝ = 0.01

The hypothesis are formulated as

H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students

against the claim

Ha: u1≠u2

i.e the GPA of night students is different from the mea GPA of day students

For two tailed test the critical value is z ≥ z∝/2= ± 2.58

The test statistic

Z= x`-u/s/√n

z= 2.84-2.55/0.45/√20

z= 0.1441

As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.

8 0

2 years ago

Determine the domain of the function, and then graph it.

Tomtit [17]

First set change the function f(x) to y so
that it would be
$y = \sqrt{x + 4} - 1$
Then set y = 0
$0 = \sqrt{x + 4} - 1$
Then solve for x

X = - 3

To graph it, just plot the point (-3,0) on the x-axis

6 0

2 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

7 0

2 years ago

4=-8p-7+9p find the solution to the equation or inequality?

andrey2020 [161]

Subtract 8p from 9p and then rewrite the equation. Then you add 7 to 4 and rewrite the equation again. Then you divide 1p by 11 and get p=11.

6 0

2 years ago