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1 year ago

14

Members of a baseball team raised $1141.75 to go to a tournament. They rented a bus for $880.50 and budgeted $23.75 per player f

or meals. Determine the number of players the team can bring to the tournament.

1 answer:

zlopas [31]1 year ago

4 0

They can bring 11 people

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A lightbulb manufacturer ships large consignments of lightbulbs to big industrial users. When the production process is function

Mashutka [201]

Answer:

a) P[ Pf | Ptd ] = 0,428

b) P[ Pf | Nd ] = 0,692

Step-by-step explanation:

We will call:

Pf Probability of functioning correctly 60 % or 0.6

Pfn Probability of malfunctioning 40 % or 0,4

PNd Probability of nondefective bulb

D₁ Probability of defective bulb when the process is correctly working

D₂ Probability of defective bulb when the process is in malfunction condition

Ptd Total probability of defective bulbs

Then applying theorem´s Bayes have

P [ A | B ] = P(A) * P [ B | A ] / P(B)

a) Probabilty of the process is functioning correctly given that we pick up a defective bulb

The total probability of defective bulbs is equal to, probabilty of defective bulbs when the process is Ok, plus probability of defective bulbs when the process is in malfunctioning condition, therefore

Ptd = (0,6)*(0,25) + (0,40)*(0,50) = 0,15 * 0,20 = 0,35

P[ Pf | Ptd ] = Pf * P [ Pd | Pf ] / Ptd = 0,6 * 0,25 / 0,35

P[ Pf | Ptd ] = 0,428

b) P [ Pf | Nd ]

P[ Pf | Nd ] = Pf * P [Nd | Pf ] / PNd

PNd = 1 - Ptd = 1 - 0,35 = 0,65

P[ Pf | Nd ] = 0,6 * 0,75 / 0,65 = 0,692

P[ Pf | Nd ] = 0,692

6 0

2 years ago

A wire 22 cm long has a mass of 374 g. What is the mass of 13 cm of this wire?

taurus [48]

Answer:

<u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

The up pinned pic is of inverse variation typed answer.. If u want word problem type answer here are the steps (EVEN IF THE STEPS ARE DIFFERENT ANSWERS REMAIN SAME)

Step-by-step explanation:

Mass of wire from 22cm = 374g

Mass of wire from 1 cm = 374÷22 = 17g

Mass of wire from 13 cm = 13×17 = <u>2</u><u>2</u><u>1</u><u> </u><u>g</u>

7 0

1 year ago

How much heat is required to raise the temperature of

Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = $T_1$ = 15°c

The final temperature = $T_2$ = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q = m × s × ( $T_2$ - $T_1$ )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q = 50 g × 2260.0 J/g × 85°c

∴ Q = 9,605,000 joule

Or, Q = 9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

If the heights of 300 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many studen

Lunna [17]

Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation

Calculate z-scores for the following random variable and determine their probabilities from standard tables.

x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088

x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912

x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587

x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413

Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36

Answer: 27

Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36

Answer: 27

Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78

Answer: 204

Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150

Answer: 150

3 0

2 years ago