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2 years ago

Which of the following numbers is prime? 12 15 7 6

Mathematics

2 answers:

vredina [299]2 years ago

5 0

Answer:

Step-by-step explanation:

A prime number is a number which has no factors other than itself and 1. In other words, this number can only be divided by itself and 1 and result in a whole number answer. All of the other choices can be divided by some other numbers and result in a whole number answer.

Factors of 12:

1, 2, 3, 4, 6, 12

Facotrs of 15:

1, 3, 5, 15

Facotrs of 7:

1, 7

Factors of 6:

1, 2, 3, 6

Kipish [7]2 years ago

3 0

Answer:

The number (7) is a prime number.

<h3>What are prime and composite numbers?</h3>

<em>Prime numbers</em> are such numbers that should have two factors, one and the number itself. An example of a prime number could be "2".

Factors of 2: 1 and 2 (Verifies the above definition)

<em>Composite numbers</em> are such numbers that have more than two factors. An example of a composite number could be 4.

Factors of 4: 1, 2 and 4 (Verifies the above definition)

<h3><u>Option A</u></h3>

Given number: 12

Let's list the factors of this number.

⇒ Factors of 12: 1, 2, 3, 4, 6, 12

The number provided is not a prime number because the number has more than two factors. Therefore, Option A is not correct.

<h3><u>Option B</u></h3>

Given number: 15

Let's list the factors of this number.

⇒ Factors of 15: 1, 3, 5, 15

The number provided is not a prime number because the number has more than two factors. Therefore, Option B is not correct.

<h3><u>Option C</u></h3>

Given number: 7

Let's list the factors of this number.

⇒ Factors of 7: 1 and 7

The number provided is a prime number because the number has exactly two factors. Therefore, Option C is correct.

<h3><u>Option D</u></h3>

Given number: 6

Let's list the factors of this number.

⇒ Factors of 6: 1, 2, 3, and 6

The number provided is not a prime number because the number has more than two factors. Therefore, Option D is not correct.

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Learn more about Prime and Composite numbers:

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Solve each equation using the quadratic formula. Find the exact solution, then approximate the solution to the nearest hundredth

Anika [276]

Answer:

$x_1=\frac{5+\sqrt{33}}{4}$ ≈2.69

$x_2=\frac{5-\sqrt{33}}{4}$ ≈-0.19

Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

1. You have that the quadratic formula is:

$x=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}$

2. To solve the quadratic equation you must substitute the values. So, you have that:

Rewrite the equation:

$2x^{2}-5x-1=0$

$a=2\\b=-5\\c=-1$

Then you have:

$x=\frac{-(-5)+/-\sqrt{(-5)^{2}-4(2)(-1)}}{2(2)}$

3. Therefore, you obtain the following result:

$x_1=\frac{5+\sqrt{33}}{4}$ ≈2.69

$x_2=\frac{5-\sqrt{33}}{4}$ ≈-0.19

4 0

3 years ago

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The volume of a cube is given by V = x3, where x is the length of an edge of the cube. The area of a square is given by A = x2,

ki77a [65]

X = 12 and A = 144 (If V = 1,728)

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3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider