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RUDIKE [14]
3 years ago
11

there are 20 students in the band .the wind pipe players make up 3/4 of the band .how my people in the band play wind pipes?

Mathematics
1 answer:
Katarina [22]3 years ago
7 0
15 people. 1/4 of the band is 5, 2/4 or half is 10, 3/4 is 15, 4/4 is all 20
You might be interested in
27 milliliters .1/10 cm is the one and ten is the mm one =
Gnesinka [82]
10 mm= 1cm
  27 mm=?
27/10*1cm=2.7cm

8 0
3 years ago
The function f(x) = x3 – 8x2 + x + 42 has zeros located at 7, –2, 3. Verify the zeros of f(x) and explain how you verified them.
xeze [42]

Answer:

  • zeros are {-2, 3, 7} as verified by graphing
  • end behavior: f(x) tends toward infinity with the same sign as x

Step-by-step explanation:

A graphing calculator makes finding or verifying the zeros of a polynomial function as simple as typing the function into the input box.

<h3>Zeros</h3>

The attachment shows the function zeros to be x ∈ {-2, 3, 7}, as required.

<h3>End behavior</h3>

The leading coefficient of this odd-degree polynomial is positive, so the value of f(x) tends toward infinity of the same sign as x when the magnitude of x tends toward infinity.

  • x → -∞; f(x) → -∞
  • x → ∞; f(x) → ∞

__

<em>Additional comment</em>

The function is entered in the graphing calculator input box in "Horner form," which is also a convenient form for hand-evaluation of the function.

We know the x^2 coefficient is the opposite of the sum of the zeros:

  -(7 +(-2) +3) = -8 . . . . x^2 coefficient

And we know the constant is the opposite of the product of the zeros:

  -(7)(-2)(3) = 42 . . . . . constant

These checks lend further confidence that the zeros are those given.

(The constant is the opposite of the product of zeros only for odd-degree polynomials. For even-degree polynomials. the constant is the product of zeros.)

3 0
2 years ago
What is the equation of the line that passes through (-2,2) and (1,-4)?
adelina 88 [10]

Answer:

y=-2x-2

Step-by-step explanation:

Slope: -4-2/1-(-2)=-6/3=-2

Equation: y=-2x+b

Substitute: 2=-2(-2)+b

2=4+b

b=-2

Therefore, y=-2x-2

I hope this helped and have a good rest of your day!

6 0
3 years ago
For which acute angles is the following statement true? The complement of the angle is smaller in measure than either of the two
Hitman42 [59]

Let the acute angle be 2x.

Its complement is 90 - 2x.

It is given that this complementary angle is less than each of the bisected angle.

Therefore, 90 - 2x < x

90 < 3x

or, x > 30.

2x > 60

Therefore, the statement is true for all angles that are greater than 60 degrees.

7 0
4 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

Compute the value of P (X ≥ 20) as follows:

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
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