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3 years ago

Is 51/17 rational or irrational?

Mathematics

2 answers:

igomit [66]3 years ago

4 0

Answer:

yes it is a rational because ia in the form of p\q

bekas [8.4K]3 years ago

3 0

Answer:

$\frac{51}{17}$ is a rational number.

Step-by-step explanation:

As, $\frac{51}{17}$

can be written in the form p/q where p is a integer and q is not equally to 0.

You might be interested in

A ternary digit is either 0, 1, or 2. how many sequences of eight ternary digits containing a single 2 and a single 1 are possib

PIT_PIT [208]

Answer:

Step-by-step explanation:

If there can be only one 1 and only one 2, the remaining digits must all be 0. The digits 1 and 2 can be anywhere in the 8 digits, and can be in either order.

There are 8 possible locations in the sequence for the 1, then 7 possible locations for the 2. The total number of possibilities is 8·7 = 56.

12000000, 21000000, 10200000, 20100000, ..., 00000012, 00000021.

8 0

3 years ago

Write the equation of the line with a slope of 1/2

Arada [10]

Answer:

y = (-1/2)x - 1

Step-by-step explanation:

y - 1 = (-1/2)(x - (-4))

y - 1 = (-1/2)(x + 4)

y - 1 = (-1/2)x -(1/2)(4)

y - 1 = (-1/2)x - 2

y = (-1/2)x -2 + 1

3 0

3 years ago

27 millimeters multiplied by 1 cm/10 mm

Basile [38]

27 millimeters multiplied by 1 cm/10 mm= 2.7 centimeters

6 0

3 years ago

4-member curling team is randomly chosen from 6 grade 11 students and 8 grade 12 students. What is the probability that the team

Ivanshal [37]

Answer:

0.5944 = 59.44% probability that the team has at least 2 grade 11 students

Step-by-step explanation:

The students are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

We have that:

6 + 8 = 14 students, which means that $N = 14$

6 grade 11 students means that $k = 6$

Teams of 4 members means that $n = 4$

What is the probability that the team has at least 2 grade 11 students?

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,14,4,6) = \frac{C_{6,0}*C_{8,4}}{C_{14,4}} = 0.0699$

$P(X = 1) = h(1,14,4,6) = \frac{C_{6,1}*C_{8,3}}{C_{14,4}} = 0.3357$

Then

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0699 + 0.3357 = 0.4056$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4056 = 0.5944$

0.5944 = 59.44% probability that the team has at least 2 grade 11 students

5 0

3 years ago

Mr .Baker has baked some muffins. If he packs them in boxes of 4, he will have 3 left over. If he packs them in boxes of 5, he w

babunello [35]

Answer:

$\boxed{43}$

Step-by-step explanation:

I think the easiest way to solve this problem is by brute force: trial and error.

We must find numbers that when divided by

4 leave 3 (4n + 3)
5 leave 3 (5n + 3)
6 leave 1 (6n + 1)

Here is a list of multiples of the integer n that satisfy the three conditions. $\begin{array}{c|ccc}\mathbf{n} & \mathbf{4n+ 3} & \mathbf{5n + 3} & \mathbf{6n + 1}\\1 & 7 & 8 & 7\\2 & 11 & 13 & 13\\3 & 15 & 18 & 19\\4 & 19 & 23 & 25\\5 & 23 & 28 & 31\\6 & 27 & 33 & 37\\7 & 31 & 38 & \mathbf{43}\\8 & 35 & \mathbf{43} & 49\\9 & 39 & 48 & 55\\10 & \mathbf{43} & 53 & 61\\\end{array}\\\text{The only number that is common to all three lists is $\mathbf{43}$.}\\\text{The smallest possible number of muffins Mr. Baker could have baked is $\boxed{\mathbf{43}}$.}$

Check:

$43 \div 4 = 10R3\\43 \div5 = 8R3\\43 \div 6 = 7R1$

OK .

6 0

3 years ago