Answer:
The answer is option C i.e; y=0.9x-45
Step-by-step explanation:
When you substitute the values of x you have to get same values as given in the table
For example: substitute x=50 in the equation y=0.9x-45
Y= 0.9(50)-45
Y= 45-45
Y=0
There ya go, Hope, this helps
Answer: z(e) = 2.07
Step-by-step explanation:
1.-The problem is about a test of proportions. As research company claims than no more 55 % of Americans regularly watch Fox News
The null hypothesis is H₀ P₀ ≤ 0.55 from 55%
And the alternative hypothesis is Hₐ Pₐ > .55
Is one tail test
2.-We have to specify significance level we assume our test will be for a significance level α = 5% or α = 0,05
3.-Calculation of z (c) = ?? and z (e) = '??
For z (c) we find in z table the value of z(c) = 1.64
For z (e) = ( P -P₀)/√p₀q₀/n z(e) = 0.05 / √(0.55*0,45)/425 z(e) = 0,05/ 0.02413 z(e) = 2.07
z(e) > z(c) threfore z(e) is in the rejection zone . We reject null hyothesis
Answer and Step-by-step explanation:
C(x) be the statement "x has a cat"
D(x) be the statement "x has a dog"
F(x) be the statement "x has a ferret".
Universe = x = all students in your class.
a) A student in your class has a cat, a dog and a ferret
= ∃x(C(x) ∧ D(x) ∧ F(x))
b) All students in your class have a cat, a dog, or a ferret = ∀x(C(x) ∨ D(x) ∨ F(x))
c) Some students in your class has a cat and a ferret but not a dog = ∃x (C(x) ∧ F(x) ∧ ¬D(x))
d) No student in this class has a cat, a dog and a ferret ¬∃x (C(x) ∧ D(x) ∧ F(x))
e) For each of the three animals, cats, dogs and ferrets, there is as student in your class who has one of the three animals. (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))