a = 30
a = 16
-a² + 46a -480 = 0
(-a + 30) (a -16) = 0
(-a)(a) + (-a)(-16) + 30(a) + 30(-16) = 0
-a² + 16a + 30a -480 = 0
-a² + 46a - 480 = 0
(-a + 30) = 0 ; (a - 16) = 0
-a = -30 ; a = 16
a = 30
To check:
a = 30
-(30)² + 46(30) - 480 = 0
-900 + 1380 - 480 = 0
480 - 480 = 0
0 = 0
a = 16
-(16)² + 46(16) - 480 = 0
-256 + 736 - 480 = 0
480 - 480 = 0
0 = 0
Answer:
The data is skewed, and the lowest number of crackers in a package was 7
Step-by-step explanation:
Hi,
First of all, since the question was incomplete due to the missing capture of the range shown on the box plot. I attached it for you so I could answer your question as well.
Taking into consideration the attached image's information, symmetric would be right down the middle, but it is not.
The image shows that it is <em>positively skewed with the lowest number being 7.</em>
Answer: 99 and 97
Step-by-step explanation: If you make a numberline and write the numbers 90 through 100. 98 is in between integers 99 and 97.
Answer:
B, C and D
Step-by-step explanation:
dilation is the only transformation that affects the size of the shape, and no transformations affect the shape of the figure.
Answer:
<h3>
ln (e^2 + 1) - (e+ 1)</h3>
Step-by-step explanation:
Given f(x) = ln and g(x) = e^x + 1 to get f(g(2))-g(f(e)), we need to first find the composite function f(g(x)) and g(f(x)).
For f(g(x));
f(g(x)) = f(e^x + 1)
substitute x for e^x + 1 in f(x)
f(g(x)) = ln (e^x + 1)
f(g(2)) = ln (e^2 + 1)
For g(f(x));
g(f(x)) = g(ln x)
substitute x for ln x in g(x)
g(f(x)) = e^lnx + 1
g(f(x)) = x+1
g(f(e)) = e+1
f(g(2))-g(f(e)) = ln (e^2 + 1) - (e+ 1)
