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2 years ago

A game has 15 balls for each of the letters B, I, N, G, and O. The table shows the results of drawing balls 1,250 times.

1 answer:

4 0

Theoretically, each letter should have the same probability of occurring since there are 15 of each. There are 5 letters that can be drawn, so there is a total of 75 balls, and each letter has a probability $\dfrac{15}{75}=\dfrac15$ of being drawn.

This means one would expect a theoretical frequency of $\dfrac{1250}5=250$ , i.e. any given letter should get drawn 250 times, which means the answer is D.

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A car and a truck left Washington, DC traveling in opposite directions. The car traveled 19 miles per hour faster than the truck

sertanlavr [38]

Truck speed be x
car travel at x+19
they travel in opposite directions
so the separation is (x+x+19) * no. of hours
1048 = (2x+19)*8
x = ((1048/8) - 19)/2
=56

truck speed is 56 miles per hour
car speed is 56+19 = 75 miles per hour

6 0

1 year ago

Read 2 more answers

A container has the shape of an open right circular cone, as shown. The height of the container is 10 cm and the diameter of the

SVETLANKA909090 [29]

A) Volume = (1/12)pi*h^3, with height = 5cm.

<span>b) You should be able to differentiate V = (1/12)pi*h^3 with respect to h, and you were given dh/dt = -0.3 cm/hr.
</span>
does that make sense?

5 0

2 years ago

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 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

Justin receives $15 and puts it into his savings account. He adds $0.25 to the account each day for a number of days, d, after t

MAVERICK [17]

Answer:

expression a

Step-by-step explanation:

The given expression is 15+0.25(d−1).

let suppose,

15 = a

0.25(d−1) = b

we get a + b

It clearly indicates the given expression is sum of two entities, we can exclude option b and option d.

Now we are left with option a and c, for that we have to evaluate the term b

b = 0.25(d−1) <u>that is the additional amount after d days</u>

Therefore, expression a is correct.

4 0

2 years ago

Read 2 more answers