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IrinaVladis [17]
4 years ago
10

What is the value of the expression? 56−(18÷34) as a fraction in simplest form

Mathematics
1 answer:
marta [7]4 years ago
7 0
The value of the expression in simplest form is: 55 8/17.

Hope this helps! :D

~PutarPotato
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4x2 + 4x =- 4<br> Find the solutions to the Quadratic equation.
VashaNatasha [74]

Answer:

x=-3

Step-by-step explanation:

(4)(2) + 4x = -4

                   multiply the 4 and 2

8 + 4x = -4

                   subtract 8 from both sides

4x = -12

                   divide both sides by 4

x = -3

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4 years ago
How many zeroes are at the end of the product?<br> 25x25x25x25x25x25x25x8x8x8
marishachu [46]

There are 9 0's in the product

6 0
3 years ago
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Hello, I need this question for today but I dont get it.. Can someone please help me?​
Bas_tet [7]

Answer:

<em>A = 32²cm</em>

Hope this helped!

Step-by-step explanation:

<h3>A = lw/2</h3>

<u>We know that the width is 8cm, and that the length of the rectangle and triangle combined is 14cm.</u>

<u />

How we find the length of the triangle itself is 14cm - 6cm, since 6cm is the rectangles length (which we know because it gives it to us at the top!)

14cm - 6cm = 8cm

<h3>Now we can plug into our formula and solve</h3>

A = 8(8)/2

A = 64/2

<h3><em>A = 32²cm</em></h3>
3 0
2 years ago
Integrate <img src="https://tex.z-dn.net/?f=%5Cint%20%5C%3A%20%5Cfrac%7Bdx%7D%7B1%20%2B%20%5Csin%28x%29%20%7D%20" id="TexFormula
Anuta_ua [19.1K]
INTEGRATION \\ \\ \\ Given \: expression \: - \: \\ \\ \int \: \frac{dx}{1 + \sin(x) } \\ \\ \int \frac{1}{(1 + \sin(x) } \times \frac{(1 - \sin(x)) }{(1 - \sin(x)) } \: dx \\ \\ \int \: \frac{(1 - \sin(x)) }{(1 - { \sin }^{2} (x))} \: dx \\ \\ \int \: \frac{(1 - \sin(x)) }{ { \cos }^{2}(x) } \: dx \\ \\ \int \: ( \frac{1}{ {cos}^{2}(x) } - \frac{ \sin(x) }{ {cos}^{2} (x)} ) \: dx \\ \\ \int( {sec}^{2} (x) - \sec(x)\tan(x) ) \: dx \\ \\ \int( {sec}^{2} (x) \: dx \: - \int \sec(x)\tan(x) ) \: dx \: \: \\ \\ \\ \: \tan(x) - \sec(x) + C \: \: \: \: \: \: \: \: \: Ans.
4 0
4 years ago
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Suppose that the lifetime of tv tubes are normally distributed with a standard deviation of 1.1 years. Suppose that exactly 20%
IrinaK [193]

Answer:

4.924 years

Step-by-step explanation:

Lets denote X the lifetime of a tv tube (In years). X has distribution N(\mu, 1.1) , with \mu unknown.

We know that P(X < 4) = 0.2. Using this data, we can find the value of \mu throught standarization.

Lets call Z = \frac{X - \mu}{1.1} the standarization of X. Z has distribution N(0,1), and its cummulative function, \phi is tabulated. The values of \phi can be found in the attached file.

P(X < 4) = P(\frac{X - \mu}{1.1} < \frac{4 - \mu}{1.1}) = P(Z < \frac{4 - \mu}{1.1}) = \phi(\frac{4 - \mu}{1.1}) = 0.2

The value q such that \phi (q) = 0.2 doesnt appear on the table. We can find it by using the symmetry of the normal density function. The opposite of q, -q must verify that \phi(-q) = 0.8 , hence -q must be equal to 0.84. Thus, q = -0.84

But this value of q should match with the number \frac{4 - \mu}{1.1} , so we have

\frac{4- \mu}{1.1} = -0.84

4 - \mu = 1.1 * (-0.84) = -0.924

\mu = 4 - (-0.924) = 4.924

Thus, the expected lifetime of TV tubes is 4.924 years.

I hope this works for you!

Download pdf
7 0
4 years ago
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