Ask question

Ask question

All categories

3 years ago

10

As part of a survey, 300 girls were asked to name their favorite sport. The results showed that 12 of the girls named bowling as

their favorite sport. What percentage of the girls in the survey choose bowling as their favorite sport?

1 answer:

Delvig [45]3 years ago

4 0

You do (12/300)x100 and it is going to give you the a answer

You might be interested in

Find the slope and y-intercept of the line.

shusha [124]

Answer:

The slope would be the gradient of the line and hence by comparison with the equation form y=mx+c where m is the gradient, the gradient of the line in the question would be 7/4.

The y intercept of the line would be where x=0. When x=0, the y intercept is thus -10.

Hence the answer is the first option

4 0

3 years ago

Read 2 more answers

Erin wants to equally divide oatmeal bars among 5 friends. If she has 12 bars, how many bars will each friend receive

Digiron [165]

Answer:

each friend will receive 2 oatmeal bars and their will be 2 leftover.

Step-by-step explanation:

6 0

3 years ago

Reiki has 3/4 of a gallon of milk. She drinks one-third of the milk. How much milk does she drink?

hammer [34]

It should be 1/4 because one third of three fourths is one fourth. I hope this helps!

4 0

3 years ago

Read 2 more answers

Please help with question :`D

olga55 [171]

Answer:

5.29 * 10^-5 is written as scientific notation.

For example:

9200 ---> 9.2 * 10³

123000 ---> 1.23 * $10^{5}$

8640000 ---> 8.64 * $10^6$

where first number/letter should be between 1 - 10.

4 0

2 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago