All categories

3 years ago

Create a list of 7 numbers with a mean of 6, a median of 6 and a range of 4.

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

8 0

Answer:

4 5 5 6 7 7 8

Step-by-step explanation:

4 5 5 6 7 7 8

For a median of 6, the 4th (middle) number must be 6.

For a range of 4, the first and last numbers must be 4 and 8, since 8 - 4 = 4.

For a mean of 6, the 7 numbers must add up to 42, since 42/7 = 6.

4 + 5 + 5 + 6 + 7 + 7 + 8 = 42.

You might be interested in

Ididmidifkjfjkelajlwkgkl5lgltwemigkktimgkgtmr

Nostrana [21]

Answer:

I aint never seen 2 pretty best friends ;)

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is 4 1/2 divided by 7 1/3 and multiplied by 1 3/5?

Luba_88 [7]

I did the work and it says 0.3683492496 and more number but i believe its 0.3 without all those other numbers

6 0

3 years ago

There are 325 children at the zoo . If 64% of children are boys and the

aleksandr82 [10.1K]

Answer:

117 Girls

Step-by-step explanation:

If 64% of the kids are boys, then 36% are girls 64% + 36% = 100%

.36 x 325 = 117 girls

Proof

.64 x 325 = 208 boys

208 + 117 = 325

4 0

3 years ago

Find x if AB = 9, BC = 2x - 5,<br> and AC = x + 9

Oxana [17]

Answer:

x = 5

Step-by-step explanation:

Using the fact that line segments AB and BC are parts of the whole line segment AC, we can write the following equation:

AB + BC = AC

Now, using the given values, we can substitute in for the equation and solve for x:

AB + BC = AC

9 + 2x - 5 = x + 9

2x + 4 = x + 9

x = 5

Thus, we have found that for these sets of equations for these line segments, our value for x should be 5.

Cheers.

5 0

3 years ago

Read 2 more answers

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

4 years ago