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3 years ago

A single 6-sided die is rolled. What is the probability of rolling a number that is not 4?

Mathematics

1 answer:

s2008m [1.1K]3 years ago

5 0

Answer:

0.5 or 3/6.. thats my guess

Step-by-step explanation:

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B. Write a part-to-part ratio using colon notation.

UNO [17]

Answer:

3:5, 4:9, 6:8 those are examples

Step-by-step explanation:

4 0

3 years ago

3x^2 - 4x = -2<br> Solve

amid [387]

Answer:

$x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$

Step-by-step explanation:

$3 {x}^{2} - 4x = - 2 \\ 3 {x}^{2} - 4x + 2 = 0 \\ equating \: it \: with \\ a {x}^{2} + bx + c = 0 \\ a = 3 \: \: b = - 4 \: \: c = 2 \\ {b}^{2} - 4ac \\ = {( - 4)}^{2} - 4 \times 3 \times 2 \\ = 16 - 24 \\ = - 8 \\ \ {b}^{2} - 4ac < 0 \\ \therefore \: given \: quadratic \: equation \: have \: \\ imaginary \: solutios. \\ \\ x = \frac{ - b \pm \sqrt{{b}^{2} - 4ac } }{2a} \\ = \frac{ - ( - 4) \pm \sqrt{ - 8} }{2 \times 3} \\ = \frac{ 4 \pm 2\sqrt{2} \: i }{2 \times 3} \\ = \frac{ 2 \pm \sqrt{2} \: i }{3} \\ \therefore \: x = \frac{ 2 - \sqrt{2} \: i }{3} \: or \: x = \frac{ 2 + \sqrt{2} \: i }{3} \: \\ \\ x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$

7 0

3 years ago

What is the range of the relation?

4vir4ik [10]

Hello!

The range is all of the y-values in a set of points. They are always ordered from least to greatest. It is written as below.

If we look at the y-values on the graph, we get -2,-3,0,and 3. If we order them, we get -3,-2,0, and 3.

This means our answer is B.

I hope this helps!

6 0

3 years ago

20 POINTS and BRAINLY!!

Karo-lina-s [1.5K]

Answer:

396

Step-by-step explanation:

3 0

3 years ago

There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac

Nataliya [291]

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$

5 0

3 years ago