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2 years ago

A dog gave birth to 9 puppies, of which 2 are brindle. What is the ratio of brindle puppies to non-brindled puppies?

Mathematics

1 answer:

Alborosie2 years ago

7 0

Answer:

2:7

Step-by-step explanation:

If 2 of the puppies are brindle that leaves 7 that are not out of the nine making the ratio 2:7.

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Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.

Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it $\vec{u}$ . This explanation uses both representations.

$\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right)$ .

$\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right)$ .

$\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right)$ .

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

$3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right)$ .

So is the case when the constant is negative:

$-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right)$ .

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

$3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right)$ .

Apply the two rules for the four vector operations.

$\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}$

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

$\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}$ .

$\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}$ .

7 0

3 years ago

A cheetah run 70 miles per hour what is the speed for feet per hour

pochemuha

Answer:

The correct answer to this question is that the cheetah can run 369,000 feet per hour. We can work this out in the following way:

One mile equals 1760 yards

In feet that is 3 x 1760 or 5,280

So Feet per hour will be 5280 x 70 or

369,600 feet per hour.

Read more on Brainly.com - brainly.com/question/134707#readmore

Step-by-step explanation:

5 0

3 years ago

Im new to this app can someone help me please and thank u so much

GaryK [48]

Answer:

$3\frac{1}{4}$

Step-by-step explanation:

we have to divide the given distance by 3.

Then we will be able to find the distance between markers.

Given distance is

$9\frac{3}{4} \\=\frac{36+3}{4}\\ =\frac{39}{4}$

Since, there are three markers,

So, we have to divide this distance by 3

So, the distance between markers is-

$\frac{\frac{39}{4}} {3} \\=\frac{39}{4} * \frac{1}{3} \\=\frac{13}{4}$

So, the distance between marker is

$\frac{13}{4}$

Now, we have to convert this into mixed number.

Then, the distance is-

$\frac{13}{4}\\ =3\frac{1}{4}$

4 0

3 years ago

Missy’s rotation maps point k (17,-12)to k (12,17) which describes the rotation?

Dennis_Churaev [7]

Answer:

This would be a 90 degree rotation in a counter clockwise fashion.

Step-by-step explanation:

You can tell this because in a 90 degree shift, the values switch terms. Also, one of the terms will change the sign.

4 0

3 years ago

Read 2 more answers

Do oddsmakers believe that teams who play at home will have home field advantage? Specifically, do oddsmakers give higher point

anygoal [31]

Complete question is;

Do oddsmakers believe that teams who play at home will have home field advantage? Specifically, do oddsmakers give higher point spreads when the favored team plays home games as compared to when the favored team plays away games?

Two samples were randomly selected from three complete National Football League seasons (1989, 1990, and 1991). The first sample consisted of 50 games, where the favored team played in a home game, while the second sample consisted of 50 games, where the favored team played in an away game. The oddsmakers’ point spreads (which are the number of points by which the favored team is predicted to beat the weaker team) were then collected.

If µ1 and µ2 represent the mean point spread for home games and away games, respectively, which of the following is the appropriate.

A) H0: μ1 - μ2 = 0

Ha: μ1 - μ2 < 0

B) H0: μ1 - μ2 = 0

Ha: μ1 < μ2

C) H0: μ1 - μ2 > 0

Ha: μ1 - μ2 = 0

D) H0: μ1 - μ2 = 0

Ha: μ1 - μ2 > 0

E) None of the above

Answer:

D) H0: μ1 - μ2 = 0

Ha: μ1 - μ2 > 0

Step-by-step explanation:

We want to find out if oddsmakers give higher point spreads when the favored team plays home games as compared to when the favored team plays away games.

Now, since µ1 and µ2 represent the mean point spread for home games and away games, respectively;

It means we want to find out if µ1 > µ2 as the alternative hypothesis.

Thus, alternative hypothesis is;

Ha: µ1 - µ2 > 0

Meanwhile null hypothesis assumes that equal point spreads are given when the favored team plays home games as well as when the favored team plays away games.

Thus, null hypothesis is;

H0: μ1 - μ2 = 0

The only correct option is D.

8 0

2 years ago