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katen-ka-za [31]
3 years ago
6

Explain Steps. Do not answer. #10 and 11Questions are in the Picture

Mathematics
1 answer:
svet-max [94.6K]3 years ago
8 0

Answer:

Step-by-step explanation:

For problem 10:

1. AE/ED=AC/CB (Since triangle ABC is similar to triangle ADE, we can determine that the ratio of AE to ED is equal to the ratio of AC to CB)

2. AE/ED=(AE+EC)/CB (Rewrite AC as the sum of the lengths forming it; This is sometimes referred to as the Partition Postulate)

3. 9/x=(9+6)/10 (Substitute the given values into this equation)

4. x=6 (Use algebra to solve for x)

For problem 11:

1. AG/AB=AE/AD (Use the same strategy as step one in problem 10, since the rectangles are similar we can create this equation)

2. AG/(AG+GB)=AE/(AE+ED) (Rewrite sides as the some of their parts)

3. 14/(14+7)=18/(18+x) (Substitute given values)

4. x=9 (Solve for x)

lmk if there are mistakes in my explanation, hope this helps :)

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3 0
3 years ago
There are 15 identical pens in your drawer, nine of which have never been used. On Monday, yourandomly choose 3 pens to take wit
DaniilM [7]

Answer: p = 0.9337

Step-by-step explanation: from the question, we have that

total number of pen (n)= 15

number of pen that has never been used=9

number of pen that has been used = 15 - 9 =6

number of pen choosing on monday = 3

total number of pen choosing on tuesday=3

note that the total number of pen is constant (15) since he returned the pen back .

probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

probability of picking a pen that has been used on tuesday = 6/15 = 2/5

probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5

on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .

we know that

probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.

to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution

p(x=r) = nCr * p^{r} * q^{n-r}

n= total number of pens=15

r = number of pen chosen on tuesday = 3

p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5

q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5

by slotting in the parameters, we have that

p(x=3) = 15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}

p(x=3) = 455 * 0.4^{3} * 0.6^{12}

p(x=3) = 455 * 0.064 * 0.002176

p(x=3) = 0.0633

thus probability that 3 of the pens has been used on tuesday. = 0.0633

probability of ball that none of the 3 pen has never being used on tuesday  = 1 - 0.0633 = 0.9337

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