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Yuki888 [10]
3 years ago
15

Four runners, Fran, Gloria, Haley, and Imani, compete on a relay team. Haley is the first runner in the relay. The other runners

can run in any order. What is the sample space showing the possible orders of the other three runners?
Mathematics
2 answers:
jeyben [28]3 years ago
8 0

Answer: The sample space showing the possible orders of the other three runners is

S=\{HFGI,HGFI,HIGF,HFIG,HGIF,HIFG\}

Step-by-step explanation:

Since we have given that

Number of runners = 4

Four runners are as follows:

Fran, Gloria, Haley, and Imani.

Since Haley is the first runner in the relay,

So, Sample space showing the possible order of the other three runners.

S=\{HFGI,HGFI,HIGF,HFIG,HGIF,HIFG\}

Hence, the sample space showing the possible orders of the other three runners is

S=\{HFGI,HGFI,HIGF,HFIG,HGIF,HIFG\}

cestrela7 [59]3 years ago
6 0
The anwser is c took the test and got it right so hopefully that helps
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Alice drove 170 miles in 2.5 hours. At this rate, how far will she drive in 4 hours? miles
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3 years ago
For the triangles ABC and MNP m∠A = m∠M, AB = 16, AC = 10, MN = 7, MP = 24. Find the ratio of the A△MNP and A△ABC .
Digiron [165]

Answer:

Ratio of area of triangles MNP and ABC is 1.05

Step-by-step explanation:

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So ratio of area of triangles = \frac{Area of MNP}{Area of ABC}

                                              = \frac{168}{160}

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3 0
2 years ago
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Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)
baherus [9]

Step 1

Given;

4x^2-y^2+24x+4y+28=0

Required; To find the center that eliminates the linear terms

Step 2

\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}

Step 3

Substitute a,d,e into the vertex form

\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\  \\  \end{gathered}

Step 4

Completing the square for -y²+4y

\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\  \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}

Step 5

Substitute a,d,e into the vertex form

\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}

Step 6

\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}

Step 7

\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}

Hence the answer is (-3,2)

4 0
1 year ago
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