Ask question

Ask question

All categories

2 years ago

14

Vector u has a magnitude of 5 units, and vector v has a magnitude of 4 units. Which of these values are possible for the magnitu

de of u + v?

1 answer:

kumpel [21]2 years ago

5 0

Answer:

Remember the triangular inequality says that if u and v are vectors then

$\lvert\lvert u + v \lvert\lvert \leq \lvert\lvert u \lvert\lvert +\lvert\lvert v\lvert\lvert$

Since the magnitude always is a nonnegative number and the magnitude of u is 5 units and the magnitude of v is 4 units ,

$0\leq \lvert\lvert u + v\lvert\lvert \leq \lvert\lvert u \lvert\lvert + \lvert\lvert v\lvert\lvert = 5+ 4 =9$

Then possibles values for the magnitude of u +v are in the interval $[0,9]$

You might be interested in

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal pla

svp [43]

Here is the correct computation of the question given.

Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Listed below are the systolic blood pressures (in mm Hg) for a sample of men aged 20-29 and for a sample of men aged 60-69.

Men aged 20-29: 117 122 129 118 131 123

Men aged 60-69: 130 153 141 125 164 139

Group of answer choices

a)

Men aged 20-29: 4.8%

Men aged 60-69: 10.6%

There is substantially more variation in blood pressures of the men aged 60-69.

b)

Men aged 20-29: 4.4%

Men aged 60-69: 8.3%

There is substantially more variation in blood pressures of the men aged 60-69.

c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

d)

Men aged 20-29: 7.6%

Men aged 60-69: 4.7%

There is more variation in blood pressures of the men aged 20-29.

Answer:

(c)

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

Step-by-step explanation:

From the given question:

The coefficient of variation can be determined by the relation:

$coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

We will need to determine the coefficient of variation both men age 20 - 29 and men age 60 -69

To start with;

The coefficient of men age 20 -29

Let's first find the mean and standard deviation before we can do that ;

SO .

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{117+122+129+118+131+123}{6}$

Mean = $\dfrac{740}{6}$

Mean = 123.33

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(117-123.33)^2+(122-123.33)^2+...+(123-123.33)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{161.3334}{5}}$

Standard deviation = $\sqrt{32.2667}$

Standard deviation = 5.68

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{5.68}{123.33}*100$

Coefficient of variation = 4.6% for men age 20 -29

For men age 60-69 now;

Mean = $\dfrac{\sum \limits^{n}_{i-1}x_i}{n}$

Mean = $\frac{ 130 + 153 + 141 + 125 + 164 + 139}{6}$

Mean = $\dfrac{852}{6}$

Mean = 142

Standard deviation $= \sqrt{\dfrac{\sum (x_i- \bar x)^2}{(n-1)} }$

Standard deviation = $\sqrt{\dfrac{(130-142)^2+(153-142)^2+...+(139-142)^2}{(6-1)} }$

Standard deviation = $\sqrt{\dfrac{1048}{5}}$

Standard deviation = $\sqrt{209.6}$

Standard deviation = 14.48

The $coefficient \ of \ variation = \dfrac{standard \ deviation}{mean}*100$

$coefficient \ of \ variation = \dfrac{14.48}{142}*100$

Coefficient of variation = 10.2% for men age 60 - 69

Thus; Option C is correct.

Men aged 20-29: 4.6%

Men aged 60-69: 10.2 %

There is substantially more variation in blood pressures of the men aged 60-69.

4 0

2 years ago

Louisa has a goal of collecting 100 pounds of dog food for a local shelter. She records how many pounds of food she collects eac

Anit [1.1K]

Answer:

28 more pouds of dog food

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for t

hoa [83]

Answer:

Since p value <0.1 accept the claim that oven I repair costs are more

Step-by-step explanation:

The data given for two types of ovens are summarised below:

Group Group One Group Two

Mean 85.7900 78.6700

SD 15.1300 17.8400

SEM 1.9533 2.3840

N 60 56

Alpha = 10%

$H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0$

(Right tailed test)

The mean of Group One minus Group Two equals 7.1200

df = 114

standard error of difference = 3.065

t = 2.3234

p value = 0.0219

If p value <0.10 reject null hypothesis

4) Since p value <0.1 accept the claim that oven I repair costs are more

3 0

2 years ago

Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .

Therefore, the observed series diverges.

5 0

2 years ago

On a road map, the locations A, B and C are collinear. Location C divides the road from location A to B, such that AC:CB = 1:2.

koban [17]

Answer:

C. (-1,-2)

Step-by-step explanation:

Since C internally divides AB in the ratio AC/CB = 1/2 = m/n where m = 1 and n = 2, we use the formula for internal division.

Let A = (x₁, y₁) = (5, 16), B = (x₂, y₂) and C = (x, y) = (3, 10)

So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

x = (mx₂ + nx₁)/(m + n)

3 = (1 × x₂ + 2 × 5)/(2 + 1)

3 = (x₂ + 10)/3

multiplying through by 3, we have

9 = x₂ + 10

x₂ = 9 - 10

x₂ = -1

y = (my₂ + ny₁)/(m + n)

10 = (1 × y₂ + 2 × 16)/(2 + 1)

10 = (x₂ + 32)/3

multiplying through by 3, we have

30 = y₂ + 32

y₂ = 30 - 32

y₂ = -2

So, the coordinates of B are (-1, -2)

3 0

2 years ago