Let
x-------> the amount of
solution
y--------> the amount of
solution
we know that
so

-------> equation A
-------> equation B
substitute equation A in equation B




find the value of y


therefore
The student need
of
solution and
of
solution
<u>the answer is</u>
A) The percent values were written incorrectly in the equation
B) The amount of 7% solution should be written as 1 – x, not x – 1.
All you do is divide by 10
there are 189 whole dimes
Answer: The probability is 1/190 = 0.005
Step-by-step explanation:
The probability of ordering two specific toppings out of 20 is:
For the first selection he can order 2 of them, peperoni or sausage, so the probability for the first selection is 2/20 = 1/10 (the number of correct options divided by the total number of options)
For the second selection we have only one option, because we assume that the other one was selected previously, here we also had a total of 19 toppings because one already was selected, the probability in this selection is 1/19.
The joint probability is equal to the product of those two probabilities:
P = (1/19)*(1/10) = 1/190 = 0.005
Since there are 22 participants all in all. The possible combinations of the two picked for practice first is,
22C2 = 231
The probability of picking one from each gender will be solved through the calculation below.
(10C1)x(12C1) / 231 = 40/77
In percentage, the answer would be approximately 52%. Thus, the answer is the first choice.
Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3.
<span>If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h. </span>
<span>Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have: </span>
<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>
<span>Thus, the area of the upper triangle is: </span>
<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>
<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>
<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>
<span>Solving with the Quadratic Formula gives: </span>
<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>
<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>
<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>
<span>I hope this helps! ^^ Brainliest Please?</span><span>
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