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3 years ago

What is 7.03 as a mixed number or fraction in simplest form

1 answer:

5 0

7.03 can be converted to 7 3/100.

.03 is the equivalent to 3 hundredths, we know this because it in the hundredths place.

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In a restaurant, the proportion of people who order coffee with their dinner is p. A simple random sample of 144 patrons of the

mote1985 [20]

<h2>Answer with explanation:</h2>

Given : In a restaurant, the proportion of people who order coffee with their dinner is p.

Sample size : n= 144

x= 120

$\hat{p}=\dfrac{x}{n}=\dfrac{120}{144}=0.83333333\approx0.8333$

The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-

Null hypothesis : $H_0: p\geq0.85$ [ it takes equality (=, ≤, ≥) ]

Alternative hypothesis : $H_1: p$ [its exactly opposite of null hypothesis]

∵Alternative hypothesis is left tailed, so the test is a left tailed test.

Test statistic : $z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

$z=\dfrac{0.83-0.85}{\sqrt{\dfrac{0.85(1-0.85)}{144}}}\\\\=-0.561232257678\approx-0.56$

Using z-vale table ,

Critical value for 0.05 significance ( left-tailed test)=-1.645

Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.

Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.

8 0

3 years ago

Jose is raising money for a school trip by selling lollipops and packs of cookies. The price of each lollipop is $1 and the pric

Ratling [72]

Answer:

For the first question, he would raise $38.

Step-by-step explanation:

1 x 12 = $12

2 x 13 = $26

12 + 26 = $38

4 0

3 years ago

Please help :( im struggling

Paha777 [63]

Answer:

a= 9/4

Step-by-step explanation:

2m^(5/2)/m^(1/4)

2m^(5/2 -1/4)

2m^(9/4)

Hope you find this helpful <3

8 0

3 years ago

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

An artist wants to frame a square painting with an area of 400 square inches . she wants

Artemon [7]

(a) $x^{2}$ = 400
(b) $\sqrt{x^2} = \sqrt{400}$
$x=20,-20$
(c) Only the positive solution makes sense because you cannot have a negative side length.
(d) Since one side of the painting is 20 inches, all four sides together will be 20*4 = 80 inches

6 0

3 years ago

What is 7.03 as a mixed number or fraction in simplest form​

What is 7.03 as a mixed number or fraction in simplest form