Ask question

Ask question

All categories

3 years ago

5

The table shows the number of minutes spent doing different exercises. The mean spent exercising was 18.2 minutes. How many minu

tes were spent doing sit ups? (URGENT HELP) (ANSWER 4/11/19)

2 answers:

Dmitry [639]3 years ago

8 0

Answer:

Step-by-step explanation:

Add all the scores you do know

8 + 10 + 38 + 20 = 76

Now put in an ex for the situps.

76 + x

There are 5 subjects

(76 + x) /5 = 18.2

The average was 18.2

Multiply both sides by 5

5*(76 + x ) / 5 = 18.2 * 5

Cancel the left combine the right.

76 + x = 91

Subtract 76 from both sides

76 - 76 + x = 91 - 76

x = 15

pentagon [3]3 years ago

4 0

Answer:

15

Step-by-step explanation:

Call the unknown time x.

The mean is the sum of the numbers divided by the number of numbers.

There are 5 numbers.

(8 + 10 + 38 + x + 20)/5 = 18.2

(x + 76)/5 = 18.2

x + 76 = 91

x = 15

You might be interested in

Which property is illustrated? 8(a + c) – 8a+ 8c

inessss [21]

Answer:

commutative property

Step-by-step explanation:

6 0

3 years ago

If you roll 2 dice, what is the probability of rolling a total of 3?

siniylev [52]

3/36 (8.333%) thats the probability

4 0

2 years ago

Solve<br><br> x + 1/2 = x - 1/2

Vladimir [108]

0 = - 1

Is the answer. I hope this helps you

7 0

3 years ago

Read 2 more answers

If Joshua borrows $300 from his older sister to buy a bike and he promises to pay the total amount plus 5% simple interest in on

xenn [34]

Answer:

you multiply 300 by 5 % and that gives you 15

Step-by-step explanation:

5 0

3 years ago

kenny6666 [7]

Answer:

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

Step-by-step explanation:

Let be $\vec u_{1} = [2,3,1]$ , $\vec u_{2} = [4,1,0]$ and $\vec u_{3} = [1, 2,k]$ , $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{3}$ if and only if:

$\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O$ (Eq. 1)

Where:

$\alpha_{1}$ , $\alpha_{2}$ , $\alpha_{3}$ - Scalar coefficients of linear combination, dimensionless.

By dividing each term by $\alpha_{3}$ :

$\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}$

$\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2}$ (Eq. 2)

$\vec O$ - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

$[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]$

$[1,2,k] = [-2\cdot\lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]$

$[0,0,0] = [-2\cdot \lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]+[-1,-2,-k]$

$[-2\cdot \lambda_{1}-4\cdot \lambda_{2}-1,-3\cdot \lambda_{1}-\lambda_{2}-2,-\lambda_{1}-k] =[0,0,0]$

The following system of linear equations is obtained:

$-2\cdot \lambda_{1}-4\cdot \lambda_{2}= 1$ (Eq. 3)

$-3\cdot \lambda_{1}-\lambda_{2}= 2$ (Eq. 4)

$-\lambda_{1}-k = 0$ (Eq. 5)

The solution of this system is:

$\lambda_{1} = -\frac{7}{10}$ , $\lambda_{2} = \frac{1}{10}$ , $k = \frac{7}{10}$

The value of the constant $k$ so that $\vec u_{3}$ is a linear combination of $\vec u_{1}$ and $\vec u_{2}$ is $\frac{7}{10}$ .

4 0

4 years ago