Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Answer:
B) f(x) = 3x² - 2x + 5
Step-by-step explanation:
To find the correct quadratic function that represents the table, you must substitute [x] into the quadratic function.
A) f(x) = 3x² + 2x - 5
B) f(x) = 3x² - 2x + 5
C) f(x) = 2x² + 3x - 5
D) f(x) = 2x² - 2x + 5
A. 3(-2)² + 2(-2) - 5 =
3(4) - 4 - 5
12 - 4 - 5
12 - 9
= 3
B. 3(-2)² - 2(-2) + 5 =
3(4) + 4 + 5
12 + 9
= 21
C. 2(-2)² + 3(-2) -5 =
2(4) - 6 - 5
8 - 11
= - 3
D. 2(-2)² + 2(-2) + 5 =
2(4) - 4 + 5
8 - 4 + 5
8 + 1
= 9
Your correct answer is: B
Hello, I Am BrotherEye
Answer: C.) 111 Only
Step-by-step explanation:
Answer:
15
Step-by-step explanation:
3375 is said to be a perfect cube because 15 x 15 x 15 is equal to 3375.
∛(15 x 15 x 15) = 15
Answer:
The claim that the scores of UT students are less than the US average is wrong
Step-by-step explanation:
Given : Sample size = 64
Standard deviation = 112
Mean = 505
Average score = 477
To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.
Solution:
Sample size = 64
n > 30
So we will use z test
Formula : 
Refer the z table for p value
p value = 0.9772
α=0.05
p value > α
So, we accept the null hypothesis
Hence The claim that the scores of UT students are less than the US average is wrong
