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Grace [21]
3 years ago
8

What product lies between 5x7/10=

Mathematics
1 answer:
Paraphin [41]3 years ago
5 0
5/1 • 7/10=
35/10=
3 5/10=
3 1/2

You might be interested in
(b) Find all 3-tuples (a,b,c) satisfying
jolli1 [7]

We have a three unknown, 4 equation homogeneous system.  These always have at least (0,0,0) as a solution.  Let's write the equations, one column at a time.

1a + 0b + 0c = 0

-1a + 1b +0c = 0

0a - 1b + c = 0

0a + 0b + -1 c = 0

We could do row reduction but these are easy enough not to bother.

Equation 1 says

a = 0

Equation 4 says

c = 0

Substituting in the two remaining,

-1(0) + 1b + 0c = 0

b = 0

0(0) - 1b + 0 = 0

b = 0

The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)

4 0
3 years ago
A random variableX= {0, 1, 2, 3, ...} has cumulative distribution function.a) Calculate the probability that 3 ≤X≤ 5.b) Find the
olchik [2.2K]

Answer:

a) P ( 3 ≤X≤ 5 ) = 0.02619

b) E(X) = 1

Step-by-step explanation:

Given:

- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:

                    F(X) = P ( X =< x) = 1 - \frac{1}{(x+1)*(x+2)}

Find:

a.Calculate the probability that 3 ≤X≤ 5

b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that

Solution:

- The CDF gives the probability of (X < x) for any value of x. So to compute the P (  3 ≤X≤ 5 ) we will set the limits.

                   F(X) = P ( 3=

- The Expected Value can be determined by sum to infinity of CDF:

                   E(X) = Σ ( 1 - F(X) )

                   E(X) = \frac{1}{(x+1)*(x+2)} = \frac{1}{(x+1)} - \frac{1}{(x+2)} \\\\= \frac{1}{(1)} - \frac{1}{(2)}\\\\= \frac{1}{(2)} - \frac{1}{(3)} \\\\=  \frac{1}{(3)} - \frac{1}{(4)}\\\\= ............................................\\\\=  \frac{1}{(n)} - \frac{1}{(n+1)}\\\\=  \frac{1}{(n+1)} - \frac{1}{(n+ 2)}

                   E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]  

                   E(X) = 1

8 0
3 years ago
A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that n = 15, p = \frac{1}{5} = 0.2

What is the probability that he answers at least 10 questions correctly?

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

0.0111% probability that he answers at least 10 questions correctly

3 0
3 years ago
I really need this thank you
a_sh-v [17]

slope intercept form: y=mx+b

y=2x-3

slope is also m

b is the y-intercept

answer:

slope is 2

y-intercept is -3

8 0
2 years ago
Write the equation x² + 6x + y² – 8y = 19 in standard form.
kirza4 [7]

Answer:

Formula used :

  • <u>x²+2xy+y²</u><u> </u><u>=</u><u> </u><u>(x+y)²</u>
  • <u>x²-2xy+y²</u><u> </u><u>=</u><u> </u><u>(x-y)²</u>

→{x}^{2}  + 6x +  {y}^{2}  - 8y = 19 \\  ({x}^{2}  + 2.3.x +  {3}^{2} ) +({y}^{2}  - 2.4.y +  {4}^{2} ) = 19 +  {3}^{2}  +  {4}^{2}  \\  {(x + 3)}^{2}  +  {(y - 4)}^{2}  = 19 + 9 + 16 \\  \boxed{ {(x + 3)}^{2}  +  {(y - 4)}^{2}  = 44}✓

  • <u>3) (x+3)²+(y-4)²=44</u> is the right answer.
8 0
2 years ago
Read 2 more answers
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