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Marizza181 [45]
3 years ago
12

Which of the following values could represent the probabilities of complementary events?

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
6 0

Answer:

I DID THE TEST Two fifths and three fifths

Step-by-step explanation:

cluponka [151]3 years ago
4 0
Two fifths and three fifths
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What is the measure of angle R?
Contact [7]

Answer:

whole triangle is 180 degrees find p with 36 and 20 then add p and q and subtract by 180

Step-by-step explanation:

I'm just helping not giving answes

7 0
3 years ago
the accumulated at 4/5 inch/hour if the snow continues at this rate for 10 hours, how much snow will accumulate
vichka [17]

Answer:

8 inches

Step-by-step explanation:

4/5 * 10/1 = 40/5 = 8inches

7 0
3 years ago
An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t
USPshnik [31]
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
3 years ago
44 +9X3- 35 x 60-20​
Roman55 [17]
Answer: -2049. Explanation: Use PEMDAS to solve the equation (parenthesis, exponents, multiplication/division, and addition/subtraction). First multiply the numbers that are being multiplied. You should get 44 + 27 - 2100 - 20. Now start from the beginning, and either add or subtract. The answer is -2049.
7 0
3 years ago
Read 2 more answers
Use the distributive property to solve 7 (x-2a-9)<br><br><br> WILL GIVE BRAINLIST TEEHEE!
koban [17]

Answer:

7x-14a-63

............

8 0
2 years ago
Read 2 more answers
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