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andriy [413]
3 years ago
15

Oscar ran the 100-yard dash in 12.41 seconds. Jesiah ran the 100-yard dash in 11.85 seconds. How many seconds faster was jesiah'

s time than oscar's time
Mathematics
1 answer:
julia-pushkina [17]3 years ago
5 0
Ok to find the difference, you have to subtract. 12.41-11.85=0.56. It was 0.56s faster.
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The function composition exists an operation " ∘ " that brings two functions f and g, and has a function h = g ∘ f such that h(x) = g(f(x)).

Let the functions be f(x) = 4x² + 1 and g(x) = x² - 3

The correct answer is (f o g)(x) = 4x⁴ - 96x + 37 and

(g o f)(x) = 16x⁴ + 8x² - 2.

<h3>What is composition function?</h3>

The function composition exists an operation " ∘ " that brings two functions f and g, and has a function h = g ∘ f such that h(x) = g(f(x)). In this operation, the function g exists used for the outcome of applying the function f to x.

Given:

f(x) = 4x² + 1 and g(x) = x² - 3

a) (f o g)(x) = f[g(x)]

f[g(x)] = 4(x² - 3)² + 1

substitute the value of g(x) in the above equation, and we get

    = 4(x⁴ - 24x + 9) + 1

simplifying the above equation

    = 4x⁴ - 96x + 36 + 1

    = 4x⁴ - 96x + 37

(f o g)(x) = 4x⁴ - 96x + 37

b) (g o f)(x) = g[f(x)]

substitute the value of g(x) in the above equation, and we get

g[f(x)] = (4x² + 1)²- 3

      = 16x⁴ + 8x² + 1 - 3

simplifying the above equation

      = 16x⁴ + 8x² - 2

(g o f)(x) = 16x⁴ + 8x² - 2.

Therefore, the correct answer is (f o g)(x) = 4x⁴ - 96x + 37 and

(g o f)(x) = 16x⁴ + 8x² - 2.

To learn more about the function refer to:

brainly.com/question/26709985

#SPJ9

7 0
1 year ago
Suppose that an automobile manufacturer designed a radically new lightweight engine and wants to recommend the grade of gasoline
yawa3891 [41]

Answer:

B) 4.07

Step-by-step explanation:

First we need to calculate the mean of all the data, which is the same as the mean of the means of each grade of gasoline:

Regular    BelowRegular   Premium   SuperPremium

39.31             36.69                38.99             40.04

39.87            40.00                40.02             39.89

39.87            41.01                  39.99             39.93

X1⁻=39.68    X2⁻= 39.23       X3⁻= 39.66    X4⁻=  39.95

Xgrand⁻ = (39.68+39.23+39.66+39.95)/4 = 39.63

Next we need to calculate the sum of squares within the group (SSW) and the sum of squares between the groups (SSB), and the respective degrees of freedom):

SSW = [ (39.31-39.68)² + (39.87-39.68)² + (39.87-39.68)² ] + [ (36.69-39.23)² + (40.00-39.23)² + (41.01-39.23)² ] + [ (38.99-39.66)² + (40.02-39.66)² + (39.99-39.66)² ] + [ (40.04-39.95)² + (39.89-39.95)² + (39.93-39.95)² ] = [0.2091] + [10.2129] + [0.6874] + [0.0121] = 11.12

SSW =  11.12

Degrees of freedom in this case is calculated by m(n-1), with m being the number of grades of gasoline (4) and n being the number of trial results for each one (3), so we would have 4(3-1) = 8 degrees of freedom

SSB = [ (39.68-39.63)² + (39.68-39.63)² + (39.68-39.63)²] + [ (39.23-39.63)² + (39.23-39.63)² + (39.23-39.63)² ] + [ (39.66-39.63)² + (39.66-39.63)² + (39.66-39.63)² ] + [ (39.95-39.63)² + (39.95-39.63)² +(39.95-39.63)² ] = [0.0075] + [0.48] + [0.0027] + [0.3072] = 0.7974

SSB =  0.80

For this case, the degrees of freedom are m-1, so we would have 4-1 = 3 degrees of freedom

Now we can establish the hypothesis for the test:

H0: μ1 = μ2 = μ3 = μ4

The null hypothesis states that the means of miles per gallon for each fuel are the same, indicating that the drade of gasoline does not make a difference, therefore our alternative hypothesis will be:

H1: the grade of gasoline does makes a difference

We will use the F statistic to test the hypothesis, which is calculated like follows:

F - statistic = (SSB/m-1) / (SSW/m(n-1)) = (0.80/3) / (11.12/8) = 0.19

We know that the level of significance we are using is α = 0.05, so to find the critical value F we need to look at some table of critical values for the F distribution for the 0.05 significance level (like the attached image). Then we just need to look fot the value that is located in the intersection between the degrees of freedom we have in the numerator (horizontal) and the denominator (vertical) of the statistic (3 and 8). That critical value is:

Fc = 4.07

3 0
2 years ago
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