Answer:
Step-by-step explanation:
Th average rate of change is the slope of the secant line that goes through those 2 values of x. Of course, each value of x also has a value of y. The coordinates for these combinations of x's and y's are:
(-1, 5) and (4, 0). We can use the slope formula to find the average rate of change of this function without having to know what the function's equation is:
![m=\frac{0-5}{4-(-1)}=\frac{-5}{5}=-1](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B0-5%7D%7B4-%28-1%29%7D%3D%5Cfrac%7B-5%7D%7B5%7D%3D-1)
So the average rate of change, aka slope, between those 2 points is -1
Answer:
598=194+8x
Step-by-step explanation:
To solve you would subtract 194 from its self and 598
598=194+8x
-194 -194
404=8x
then you would divide 8 from it's self and 404
which would make it 50.5
(but you cant buy half a package of spoons so realistically it'd be 50 or 51.
hope this helped :)
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
Answer:
72![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Step-by-step explanation:
The area (A) of a rhombus is calculated as
A =
× d₁ × d₂ (d₁ and d₂ are the diagonals )
The diagonals bisect each other at right angles
d₁ = 2 × 6 = 12
Use the tangent ration in the upper left right triangle and the exact value
tan60° = ![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
tan60° =
=
=
( multiply both sides by 6 )
opp = 6
, then
d₂ = 2 × 6
= 12![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Thus
A =
× 12 × 12
= 6 × 12
= 72![\sqrt{3}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D)
Answer:
Jada should have multiplied both sides of the equation by 108.
Step-by-step explanation:
The question is incomplete. Find the complete question in the comment section.
Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:
Step 1: Multiply both sides of the equation by -9/4 as shown:
-4/9 × -9/4 = x/108 × -9/4
-36/-36 = -9x/432
1 = -9x/432
1 = -x/48
Cross multiplying
48 = -x
x = -48
It can also be solved like this:
Given -4/9 = x/108
Multiply both sides by 108 to have:
-4/9 * 108 = x/108 * 108
-4/9 * 108 = 108x/108
-432/9 = x
x = -48
Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.