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Volgvan
3 years ago
15

X-4y=-17 and y=4x +8 solve by graphing please and show the solution

Mathematics
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

The answer to your question is (-1, 4) The lines cross in one point so they have only one solution.

Step-by-step explanation:

Data

Equation 1     x - 4y = -17

Equation 2    y = 4x + 8

Solve for y

Equation 1          y = -x/-4 - 17/-4

                          y = x/4 + 17/4

Equation 2         y = 4x + 8

See the graph below

These lines cross in point (-1, 4), so that is the only one solution.

If the lines were the same line they would have infinite solutions

                   

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Suppose you toss a fair coin 10 times, let X denote the number of heads. (a) What is the probability that X=5? (b) What is the p
zubka84 [21]

Answer:  The required answers are

(a) 0.25,    (b) 0.62,    (c) 6.

Step-by-step explanation:  Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.

(a) The probability of getting 5 heads is given by

P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.

(b) The probability of getting 5 or more than 5 heads is

P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

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