The integers (4, 5) do not have real zero.
Knowing what zeros represent can assist us in determining when and how to locate the zeros of functions given their expressions and a function's graph. The value of x when the function itself reaches zero is typically referred to as a function's zero.
A function's zero can take many different forms, but as long as they have a y-value of zero, we will consider them to be the function's zero.
Given Expression
f(x) = x³ + 9x² + 8x - 5
to find which is not a real zero,
condition of real zero is for any function f(a , b) if f(a).f(b) < 0 the function have at least a zero.
1: (-8, -7)
f(-8).f(-7) = [(-8)³ + 9(-8)² + 8(-8) - 5][(-7³) + 9(-7)² + 8(-7) - 5]
f(-8).f(-7) = (-5)(37)
f(-8).f(-7) = -185 < 0 points have at least a zero
2: (4, 5)
f(4).f(5) = [(4)³ + 9(4)² + 8(4) - 5][(5³) + 9(5)² + 8(5) - 5]
f(4).f(5) = 235 x 385
f(4).f(5) = 94,475 > 0
points do not have any zeros
3: (0, 1)
f(0).f(1) = [(0)³ + 9(0)² + 8(0) - 5][(1³) + 9(1)² + 8(1) - 5]
f(0).f(1) = -5 x 13
f(0).f(1) = -65 < 0
points have a zero
4: (–2, –1)
f(-2).f(-1) = [(-2)³ + 9(-2)² + 8(-2) - 5][(-1³) + 9(-1)² + 8(-1) - 5]
f(-2).f(-1) = 7 x (-5)
f(-2).f(-1) = -35 < 0
points have a zero
Hence only point (4, 5) do not have a zero.
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Answer:
Option B (4,5)
Step-by-step explanation:
Freshman 16 22
Sophomore 24 15
Junior 17 20
Senior 25 19
Super Senior 8 5
What is the distribution of the test statistic under the null hypothesis
Answer:
There is no relationship between your year in school and having a job.
Step-by-step explanation:
In this instance, the chi sq test need to be performed.
Chi sq is used to determine if there is a significant relationship between two categorical variables.
The two variables here are year in school and employment status.
The two variables are independent(no relationship exists)
This implies that there is null hypothesis
Therefore, the Null Hypothesis is
There is no relationship between your year in school and having a job.
Answer:
Reject the null hypothesis.
Step-by-step explanation:
n1 = 16
n2 = 21
S.V1 = 5.8
S.V2 = 2.4
= 0.05
= Population Variance 1 ≤ Population Variance 2
= Population Variance 1 > Population Variance 2
Test statistic value = 5.8 / 2.4 = 2.417
Degrees of freedom is n - 1
15 and 20
Critical value is = 2.2
2.417 > 2.2
we reject the null hypothesis as the critical value is greater than the test statistic.
Answer:
Step-by-step explanation:
Total number of tickets sold = 3388
Total number of coach tickets = 3069
Total number of first-class tickets = Total number of tickets sold - Total number of coach tickets
= 3388 - 3069
= 319
Ratio of the number of first-class tickets to the total number of tickets = 319:3388
Answer:
Step-by-step explanation:
Given, Total no. of tickets sold = 3388
Total no. of coach tickets = 3069
Then, No. of first class ticket:
= 3388 - 3069
= 319
We need to find the ratio of first-class tickets to the total number of tickets: 319:3388
Answer:
Step-by-step explanation:
Assuming the order in which he answers the questions matter the answer is the number of permutations of 8 in 10.
This is 10! / (10-8)!
= 1,814.400.
If the order does not matter then the answer is the number of combinations of 8 from 10:
This is 10!/8!*2!
= 45.
The cost of 1 hot dog is $1.75 and the cost of 1 hamburger is $2.
Let x be the cost of one hot dog and y be the value of one hamburger. we will set up a system of linear equations to solve for x and y:
From the first statement, we recognize that:
7x + 2y = 16.25 ----(1)
From the second statement, we know that:
2x + 7y = 17.50 ----(2)
We will use both substitution or elimination method to solve this system of equations. let's use elimination method right here. we are able to multiply equation (1) through 7 and equation (2) by 2 to eliminate y:
49x + 14y = 113.75 ----(3)
4x + 14y = 35 ----(4)
Subtracting equation (4) from equation (3), we get:
45x = 78.75
Dividing each sides by using 45, we get:
x = 1.75
Substituting x = 1.75 into equation (1), we can solve for y:
7(1.75) + 2y = 16.25
12.25 + 2y = 16.25
2y = 4
y = 2
Consequently, the cost of 1 hot dog is $1.75 and the cost of 1 hamburger is $2.
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