A) y = 2x – 7 and f(x) = 7 – 2xIncorrect. These equations look similar but are not the same. The first has a slope of 2 and a y-intercept of −7. The second function has a slope of −2 and a y-intercept of 7. It slopes in the opposite direction. They do not produce the same graph, so they are not the same function. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5. B) 3x = y – 2 and f(x) = 3x – 2Incorrect. These equations represent two different functions. If you rewrite the first equation in terms of y, you’ll find the equation of the function is y = 3x + 2. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5. C) f(x) = 3x2 + 5 and y = 3x2 + 5Correct. The expressions that follow f(x) = and y = are the same, so these are two different ways to write the same function: f(x) = 3x2 + 5 and y = 3x2 + 5. D) None of the aboveIncorrect. Look at the expressions that follow f(x) = and y =. If the expressions are the same, then the equations represent the same exact function. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5.
The answer for the exercise is the third option, which is: Hexagon.
The explanation is shown below:
As you can see in the figure attached, the cross section is a polygon of six sides and six angles. Therefore, it has six vertexes. In geometry, this type of polygon is known as "Hexagon".
We know there’s 660 boxes and each box called for 3 eggs, therefore:
660=3X
when divide both sides by 3 to isolate the x, which will tell us the amount of eggs used…
660/3 = 220
Answer:
C
Step-by-step explanation:
x + 6 not hard
Answer:
c) Is not a property (hence (d) is not either)
Step-by-step explanation:
Remember that the chi square distribution with k degrees of freedom has this formula
Where N₁ , N₂m .... 
are independent random variables with standard normal distribution. Since it is a sum of squares, then the chi square distribution cant take negative values, thus (c) is not true as property. Therefore, (d) cant be true either.
Since the chi square is a sum of squares of a symmetrical random variable, it is skewed to the right (values with big absolute value, either positive or negative, will represent a big weight for the graph that is not compensated with values near 0). This shows that (a) is true
The more degrees of freedom the chi square has, the less skewed to the right it is, up to the point of being almost symmetrical for high values of k. In fact, the Central Limit Theorem states that a chi sqare with n degrees of freedom, with n big, will have a distribution approximate to a Normal distribution, therefore, it is not very skewed for high values of n. As a conclusion, the shape of the distribution changes when the degrees of freedom increase, because the distribution is more symmetrical the higher the degrees of freedom are. Thus, (b) is true.
