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2 years ago

Which expression shows the result of applying the distributive property to 3(15x−17) ?

Mathematics

2 answers:

Olin [163]2 years ago

8 0

Answer:

$\frac{3}{5}x-\frac{3}{7}$

Step-by-step explanation:

we have

$3(\frac{1}{5}x-\frac{1}{7})$

Applying the distributive property

$3(\frac{1}{5}x-\frac{1}{7})=3(\frac{1}{5}x)-3(\frac{1}{7})\\ \\=\frac{3}{5}x-\frac{3}{7}$

Mariulka [41]2 years ago

6 1

You just multiply 3 by both 15x and -17. None of these choices are correct. The only correct application would be A.

IDontLikeYou101

2 years ago

ITS NOT A ITS B DUMBSS

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Answer:

Step-by-step explanation:

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2 years ago

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3 years ago

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2 years ago

Read 2 more answers

Find all exact solutions on [0, 2π). (Enter your answers as a comma-separated list.)

nalin [4]

Answer:

$0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi$

Step-by-step explanation:

Given the equation:

$sin(x)-2sin(x) tan(x)=0$

To find:

The solutions of given equation in the range [0, 2 $\pi$ ) i.e. 0 can be in the answer but $2\pi$ can not be there in the answer.

Solution:

Taking $tan(x)$ common, we get:

$tan(x) (1-2sin(x))=0$

The solutions can be given as:

$tan(x) = 0$ OR $1-2sin(x) = 0$

Let us solve both the equations one by one:

First equation:

$tan(x) = 0\\ \Rightarrow x=0, \pi$

Second equation:

$1-2sin(x) = 0\\\Rightarrow 2sinx=1\\\Rightarrow sinx=\dfrac{1}{2}\\\Rightarrow x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Therefore, the answers as a comma separated list are:

$0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi$

7 0

2 years ago

Given the following observed and expected data (total of 1000), using chi-squared calculate a p-value that corresponds with this

finlep [7]

Answer:

The answer is " $0.90>p>0.75.$ "

Step-by-step explanation:

$\text{Cinnabar vestigial} \ \ \ \ \ \ \ \ \ \ \ 384 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ -6 36 \ \ \ \ \ \ \ \ \ \ \ 0.092308\\\\$

$roof \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 408 \ \ \ \ \ \ \ \ \ \ \ 390 \ \ \ \ \ \ \ \ \ \ \ 18 \ \ \ \ \ \ \ \ \ \ \ 324 \ \ \ \ \ \ \ \ \ \ \ 0.830769\\\\\text{Cinnabar vestigial roof} \ \ \ \ \ \ \ \ \ \ \ \ \ 63 \ \ \ \ \ \ \ \ \ \ \ \ \ 70\ \ \ \ \ \ \ \ \ \ \ \ -7 \ \ \ \ \ \ \ \ \ \ \ 49 \ \ \ \ \ \ \ \ \ \ \ 0.7\\\\\text{wild type} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 \ \ \ \ \ \ \ \ \ \ \ 70 \ \ \ \ \ \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 0.057143\\\\$

$vestigial \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.257143\\\\$

$\text{Cinnabar roof} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 34\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 35 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.028571\\\\\text{roof vestigial} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.2\\\\$

$\text{cinnabar} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8 \\\\$

$Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.965934$

Eight phenotypes were present.

Df is provided also by a number of phenotypes -1 The degree of freedom

$\to df = 8-1= 7$

For p-value 0,9, Chi-square is 2.83;

The p-value of 0.75 is 4.5. Chi-square

Chi-sqaure value is observed at 2.965.

That means 0.90>p-value>0.75.

7 0

3 years ago