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murzikaleks [220]

3 years ago

9

On the last day of a Shakespeare class, an English teacher asked her students which play they liked most. Out of the 18 students

who have submitted responses so far, 6 liked Much Ado About Nothing best. What is the experimental probability that the next student to respond likes Much Ado About Nothing best?
Write your answer as a fraction or whole number.

1 answer:

Vika [28.1K]3 years ago

7 0

Answer:

1/3

Step-by-step explanation:

There were 18 students and six liked Much Ado About Nothing. So, you could write this as 6/18 which simplifies to 1/3.

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The perimeter of a rectangle is 72 in. The base is 3 times the height. Find the area of the rectangle

Dovator [93]

Answer:

243 in²

Step-by-step explanation:

Step 1. Calculate the<em> base and height </em>of the rectangle

We have two conditions:

(1) 2b+ 2h = 72 in

(2) b = 3h Substitute in (1)

2(3h) + 2h = 72 Remove parentheses

6h + 2h = 72 Combine like terms

8h = 72 Divide by 6

h = 9 in Substitute in(2)

b = 3 × 9

b = 27 in

Step 2. Calculate the area of the rectangle

A = bh

A = 27 × 9

A = 243 in²

The area of the rectangle is 243 in².

7 0

3 years ago

The price that a company charged for a basketball hoop is given by the equation 50-5x^2 where x is the number of hoops that are

Debora [2.8K]

Given:
Profit : 15,000,000
Cost: 30 per basketball hoop
production: 1 million hoops
price: 50 - 5x²

Profit = Sales - Cost
15,000,000 = sales - 30(1,000,000)
15,000,000 + 30,000,000 = sales
45,000,000 = sales

45,000,000 / 1,000,000 = 45 sales price.

3 0

3 years ago

Read 2 more answers

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

<u>These are the steps to find the Taylor series for the function</u> $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

<u>To find the first three non-zero terms you need to replace n=3 into the sum</u>

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago

The following data on average daily hotel room rate and amount spent on entertainment (The Wall Street Journal, August 18, 2011)

Zielflug [23.3K]

Answer:

a. Predicted Amount = $109.46

b. Confidence Interval = (94.84,124.08)

c. Interval = (110.6883,188.8517)

Step-by-step explanation:

Given

ŷ = 17.49 + 1.0334x.

SSE = 1541.4

a.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46 --- Approximated

Predicted Amount = $109.46

b.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

s = √220.2

s = 14.839

Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148 -- 161 -- 43-- 1849

96 || 105|| -9 || 81

91 ||101 || -14 || 196

110 || 142 || 5 || 25

90 || 100 || -15 || 225

102 || ||120 ||-3|| 9

136 || 167 ||31 ||961

90 || 140 ||-15 ||225

82 || 98 ||-23 || 529

Sum 945 || 1134|| 0 ||4100

Mean (x) = 945/9 = 105

∑(x - (mean(x))² = 4100

α = 1 - 95% = 5%

α/2 = 2.5% = 0.025

tα,df = n − 2 = t0.025,7 =2.365

Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100

Confidence Interval = (109.46 ± 14.62)

Confidence Interval = (94.84,124.08)

c.

ŷ = 17.49 + 1.0334(128)

ŷ = 149.7652

ŷ = 149.77

Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100

Interval = 149.77 ± 39.0817

Interval = (110.6883,188.8517)

3 0

3 years ago

As an estimate we are told 5 miles is 8 km, what is 62.5 miles into km

Sati [7]

Answer:

≈100.5

Step-by-step explanation:

hope this helps

6 0

2 years ago