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algol [13]
3 years ago
9

5/8 of the 64 musicians in a music contest are guitarists. Some of the guitarists play jazz solos, and the rest play classical s

olos. The ratio of the number of guitarists in the contest is 1:4. How many guitarists play classical solos in the contest?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
7 0
First, find the number of guitarists. 5/8*64=40.

Now, create an equation. x + 4x = 40
5x = 40
x= 8

There are 8 jazz solos and 32 classical solos.
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Katen [24]
You don't even have to actually do the division to find the correct answer.

The leading term of q(x) is x³/x² = x, so only the 1st and 3rd selections are appropriate.

The remainder r(x)/b(x) will have the divisor as b(x), so only ...
.. the 1st selection is appropriate.
7 0
3 years ago
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If g(x) = 4x + 1 and g(x) = -11, find x.
podryga [215]

Answer:

x = - 3

Step-by-step explanation:

g(x) = 4x + 1...... (1)

g(x) = -11.... (2)

From equations (1) & (2)

-11 = 4x + 1

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x = - 3

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2 years ago
Abe,a plumber, charges $50 per hour for making an in-home visit. The Plumbing Service $65 per hour but no set free for a visit.
ArbitrLikvidat [17]

Answer:

Plumber: 13 hours

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Step-by-step explanation:

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5 0
3 years ago
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Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

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7 0
3 years ago
Can you please help me out
Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}

Hence, the best option is G.

5 0
1 year ago
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