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faltersainse [42]
4 years ago
9

How do equivalent fractions help when adding mixed numbers

Mathematics
1 answer:
prisoha [69]4 years ago
5 0
You do the butterfly method multiply across and add or subtract. multiply the denominators to get the denominators
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Solve the equation w+i = s/c for c
trasher [3.6K]

Answer:

all work is shown and pictured

4 0
3 years ago
-5 = .25d +3<br><br> Help me
Natalka [10]

Answer:

d =  - 32

Step-by-step explanation:

- 5 = 0.25d + 3 \\ add \: 5 \: from \: both \: sides \\  \\  - 0.25 = 8 \\ divide \: both \: sides \: by \:  - 0.25 \\  \\ d =  - 32

7 0
3 years ago
Mjnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
Radda [10]

I don't know what this is supposed to mean, but thanks for the free points! ;)

7 0
4 years ago
Gummi bears come in twelve flavors, and you have one of each flavor. Suppose you split your gummi bears among three people (Adam
aliina [53]

Step-by-step explanation:

From the given gummi bear, the chance that Adam is selected for any draw is 1/3 as well as the chance he is not selected at any draw is 2/3.

a). The probability of Adam getting exactly three gummi bears = P(Adam gets selected at 3 draws and not selected at the remaining 9 draws)

= $ (\frac{1}{3})^3 (\frac{2}{3})^9 = \frac{2^9}{3^{12}} $

Now, the 3 draws where Adam gets selected can be any 3 out of 12 draws in $ {\overset{12}C}_3 $ = 220

Thus, probability of Adam getting three gummi bears = $ 220 \times \frac{2^9}{3^{12}} $

                                                                                                = 0.21186

b). Probability that Adam will get the three gummi bears given each person will received at the most 1 gummi bear

= P(of the remaining 9 draws after assigning one gummi bear to each one, Adam gets selected at 2 draws and not selected at 7 draws) = $ {\overset{9}C}_2 (\frac{1}{3})^2 (\frac{2}{3})^7 $

                                                                                               = 0.23411

c). Let X = Number of the gummi bears which Adam will get. Then, X = number of draws out of 12 draws Adam gets selected and X ~ B(12, 1/3). So,  Adam will get gummi bears= mean of B(12, 1/3) = 12 x (1/3) = 4

d). Let Y = Number of the gummi bears that Adam will get, given each person will received at the most 1 gummi bear Then, Y = number of draws Adam gets selected in the remaining 9 draws and Y ~ B(9, 1/3). So, the expected number of bears that Adam gets given each person received at least 1 gummi bear

= 1 + mean of B(9, 1/3) = 4

e). When every one gets atleast one gummi bear, the new sample size will be 9 and so we can say that there is a reduction in variance.

8 0
3 years ago
Which of the following are not polynomials?
PolarNik [594]
First, ask yourself:  What is a polynomial?  It's an algebraic expression in which all powers of x are zero or positive integers.
Eliminate E because -2 is not a positive integer exponent.
Accept D because all of the powers of x are positive integers:  {3, 2, 1, 0}
Eliminate C because -3 is not a positive integer power
Eliminate A and B because of the fractional or negative exponents shown.

Only D is a polynomial.

6 0
4 years ago
Read 2 more answers
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