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Setler [38]
2 years ago
5

Calculate the lower quartile of the data set for the blood glucose levels of 10 individuals whose blood glucose readings were as

follows: 88, 97, 101, 104, 104, 107, 109, 117, 121, and 147
Mathematics
1 answer:
Vsevolod [243]2 years ago
4 0

Answer:

The correct answer is 101.

Step-by-step explanation:

Lower quartile is defined as the mid value of the first half of the data set when the data set is arranged in ascending order. Lower quartile in other words separates the lowest 25% from the highest 75% of the data.

The lower quartile of the data set for the blood glucose levels of 10 individuals is given by the mid value of the first five terms 88, 97, 101, 104 and 104.

Thus the mid value of this first half \frac{5+1}{2}  th term is 101. Thus the lower quartile of the data set is 101.

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Simplify the following 2(×+8k)+2k=​
omeli [17]

Answer:

2x + 18k = X= -9k

Step-by-step explanation:

you have to solve the rational equation by combining expressions and isolating the variable x.

5 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
To show that the triangles are congruent, triangle ABC could be transformed by
tekilochka [14]

Answer:

D a reflection over the x axis by a reflection over the y axis

Step-by-step explanation: if u put the triangle on the button left up then move it to the upper right it’s congruent.

3 0
3 years ago
Read 2 more answers
Ill give you brainliest and 5 stars help plz
Sloan [31]

Answer:

D

Step-by-step explanation:

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3 years ago
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Goryan [66]

Answer:

(TAKE THE MINUS OUT)

Step-by-step explanation:

The first thing he has to do is take that "minus" sign through the parentheses containing the second polynomial. Some students find it helpful to put a " 1 " in front of the parentheses, to help them keep track of the minus sign. Here's what the subtraction looks like, when working horizontally.

6 0
3 years ago
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