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cricket20 [7]
3 years ago
13

Select all the true sentences!!!

Mathematics
2 answers:
pishuonlain [190]3 years ago
6 0

Answer:

only the 1st is true

Step-by-step explanation:

the rest are wronf

ludmilkaskok [199]3 years ago
6 0
First is correct

...............
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c or d but i think d

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Maurinko [17]
How don’t you understand that
8 0
2 years ago
The distribution of individual incomes in the United States is strongly skewed to the right. In 2013, the mean and median income
vesna_86 [32]

Answer:

The mean would be $322,343 and the median would be $196,723.

Step-by-step explanation:

Since the distribution of individual incomes is skewed to the right, it means that the distribution has a long right tail.

Drawing a distribution with this characteristic, we can see how the majority of the data falls into the left side of the graphic, meaning that a lot of people receive less income. Following this reasoning, the mean which is the amount of the data (in this case individual income) divided by the amount of people, would be the higher number, meaning that the few people who earn more money would influence in making this number higher.

Following this reasoning, the median (which is not influenced by this difference) would be the less high number.

5 0
4 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
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