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2 years ago

Someone help me pls ill give brainlist to the correct one

Mathematics

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

the answer I got was

$\frac{6}{10} or \: \frac{3}{5}$

Step-by-step explanation:

so I included a picture of my work

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Please I need help ASAP

Vikki [24]

Answer:

{y,x}={-2,8}

Step-by-step explanation:

3 0

2 years ago

What time will Jesus come back to the Earth? Why? May you please provide the Year, date, time of day(hours and minutes). You can

Mashutka [201]

Answer:

We don't know when Jesus will come back.

Step-by-step explanation:

The bible doesnt specify

5 0

2 years ago

Fine lowest common values for x and y variables of A[4x - 7y = 2] <br> B[3x - 3y = 6]

OLEGan [10]

3x -3y = 6

or, 3(x -y)= 6

or, x-y =6/3

or, x-y = 2

or, x=2+y........eqn(i)

4x - 7y = 2

substituting the value of x from eqn(i)

or, 4(2+y) -7y = 2

or, 8 + 4y -7y = 2

or, -3y = 2-8

or, y = -6/-3

■ y=2

so in eqn(i)

x= 2+y

x= 2+2

■x = 4

3 0

2 years ago

Nadia is ordering cheesecake at a restaurant, and the server tells her that she can have up to five toppings: caramel, whipped c

stealth61 [152]

Answer:

4/5

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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.

3 0

3 years ago