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3 years ago

How would you draw a picture to find 3 ÷ 1/2

1 answer:

5 0

You draw 3 rectangles and you split them into halves or put a line in between. Then, count 1 of every line in each of the boxes. You should get 3/2 because it's out of a half and that simplified is 1 and 1/2.

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Equivalent Fractions

TEA [102]

Answer:

What is the question????

5 0

3 years ago

4x-5=2x+7 midpoint segmen

Bogdan [553]

Answer:

x = 6

Step-by-step explanation:

The way to solve this is by subtracting 2x from both sides. Now we have

2x-5=7

Then we add 5 from both sides. Now we have

2x=12

We do this because we want X by itself

Then we divide by 2 on both sides and now we have

x = 6

If you have any questions feel free to ask in the comments.

7 0

3 years ago

9. (3x + 4y)² – (3z) 2square<br>

rodikova [14]

Solution: (3x + 4y - 3z) x (3x + 4y + 3z)

5 0

3 years ago

Read 2 more answers

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

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

which converges everywhere. Then by substitution,

$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

$-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)$

c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago

According to the "January theory," if the stock market is up for the month of January, it will be up for the year. If it is down

PSYCHO15rus [73]

Answer:

The required probability is 0.031918.

Step-by-step explanation:

Consider the provided information.

It is given that According to an article in The Wall Street Journal, this theory held for 22 out of the last 34 years.

Therefore n=34

The probability it is either up or down is 0.5.

Thus the value of p =0.5 and q = 0.5

Now calculate the probability

Let X is the number of years that theory held.

X has a binomial distribution with n=34 and p=0.5

$P(X=22)=\binom{34}{22}(0.5)^{22}(0.5)^{12}\\P(X=22)=\frac{34!}{22!\times 12!}(0.5)^{22}(0.5)^{12}\\P(X=22)=548354040\times 5.8207660913\times 10^{11}\\P(X=22)=0.0319184060208\\P(X=22)\approx0.031918$

Hence, the required probability is 0.031918.

8 0

3 years ago