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svlad2 [7]
4 years ago
7

Two equations are shown:

Mathematics
2 answers:
myrzilka [38]4 years ago
8 0
Both are linear because they're on the format of y=mx+b. m= slope and b= the y-intercept
Anit [1.1K]4 years ago
8 0
<span>the answer is "Both are nonlinear. "
</span>
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What fraction of this model in simplest form is shaded (fraction is 2/10)
rodikova [14]
I don’t really understand ur question but the fraction 2/10 in simplest form is 1/5
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3 years ago
A baseball team sells tickets for two games. The ratio of sold tickets to unsold tickets for the first game was 7:3. For the sec
balandron [24]
2:40, because unsold tickets were true? or no. ask yourself that
6 0
3 years ago
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A lender estimates the closing costs on a home loan of $90,000 as listed below. Closing Cost Charge Loan origination $180 Title
AURORKA [14]

Answer:

The answer would be,

a. The lender made a very good estimate; it was within 0.25% of the actual closing costs.

Step-by-step explanation:


6 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
4 years ago
What’s the greatest common factor of 22 99
Elza [17]

The greatest common factor of 22 and 99 would be 11. :)

7 0
4 years ago
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