Work out the value of z, given that x+y=42 and x/13=y/8=z/9.
1 answer:
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First, "graph" means graph of the given function f(x). It is not necessary to include the graph with the question because at this level, either the user can "see" the graph mentally, or use technology (calculator or software) to graph the function.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.
(a) The Leading coefficient test for polynomial functions
The test is as follows. The sign of the leading coefficient (LC, coefficient of the highest power of the variable) together with the degree (odd or even) of the polynomial function tells us the end behaviour of the function according to the following table:
LC/degree Even Odd
Positive +inf/+inf -inf/+inf
Negative -inf/-inf +inf/-inf
It can be easily memorized by comparing with a line or a parabola
y=x (positive leading coefficient, odd degree) goes from -inf to +inf
y=x^2 (positive leading coefficient, even degree) goes from +inf to +inf.
Any negative leading coefficient will just invert the signs.
For the given polynomial function
f(x) = x^2(x+2) = x^3+x^2
The leading coefficient is +1 (positive), and degree is three (odd)
So it behaves the same way as a straight line y=x, so the end behviours are
{-inf, +inf} as x-> {-inf, +inf} respectively.
(b) x-intercepts
Since f(x)=x^2(x+2), the x-intercepts are x=0 (multiplicity 2) and x=-2.
=>
From the left, graph is increasing and crosses x=-2 once from negative to positive, after which graph is decreasing. At x=0, because of the root with multiplicity 2, graph touches the origin (0,0) and turns back towards +inf.
(c) the y-intercept
from the polynomial f(x)=x^2(x+2)=x^3+x^2 = x^3+x^2+0,
we see that the y-intercept =0 because the "initial value" or the constant term is zero.
<h2>Answer with explanation:</h2>
Given : In a restaurant, the proportion of people who order coffee with their dinner is p.
Sample size : n= 144
x= 120
The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-
Null hypothesis : [ it takes equality (=, ≤, ≥) ]
Alternative hypothesis : [its exactly opposite of null hypothesis]
∵Alternative hypothesis is left tailed, so the test is a left tailed test.
Test statistic :
Using z-vale table ,
Critical value for 0.05 significance ( left-tailed test)=-1.645
Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.
Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.