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vova2212 [387]
3 years ago
8

2 over 3 divided by 5

Mathematics
1 answer:
Yuliya22 [10]3 years ago
6 0
2/3 divide by 5 is 2/15

Remember keep change flip

2/3 times 1/5 is 2/15

Hope this helps!
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The transformations to the parent function of a quadratic equation are given below. write an equation of the new function in ver
musickatia [10]

Answer:

<h2>11.f(x)=^2-3</h2>

Step-by-step explanation:

<h2>12.And slide everything five units down. So a over here that's negative three negative one. So if we slide at five you just down we go one two three four five and get here.</h2>

Hope it helps

3 0
2 years ago
-k^2-(4k-9n)+6n; k= -3 and n= -2
dalvyx [7]

Answer:

\displaystyle -27

Step-by-step explanation:

\displaystyle -[-3]^2 - [4(-3) - 9(-2)] + 6[-2] \hookrightarrow -9 - [-12 + 18] - 12 \hookrightarrow -9 - 6 - 12 \\ \\ \boxed{-27}

I am joyous to assist you at any time.

6 0
3 years ago
In professor shannon's economics class there are 12 male students. females represent 60% of the class total. how many students d
Mekhanik [1.2K]
He has 30 students in his class because 40% of the class is boys and there is 2.5 40% in 100% so we do 12 x 2.5

6 0
3 years ago
Read 2 more answers
What information did Sebastian use in his deduction?
Dmitry_Shevchenko [17]

Answer:

D. The table shows two x-intercepts and it changes from increasing to decreasing.

Step-by-step explanation:

A logarithmic function isn't either of these. This is the correct answer of ed-genuity.

5 0
3 years ago
Read 2 more answers
Let X be a random variable with CDF given byFX(t) =0 for t &lt; 1,1 /2 for ?1 t &lt; 11/ 2 t for 1 t &lt; 21 for t 2Calculate E[
nydimaria [60]

Answer:

\mathbf{E(X) = \dfrac{3}{4}}

Step-by-step explanation:

From the given data, the cumulative distribution function of a random variable can be represented as:

F_X(t)  =\left\{ \begin{array}{c}0........... t

The objective is to estimate E(X), to do that, let's first evaluate the probability density function by differentiating the cumulative distribution function from above.

f_X(x) =\left \{ {{\dfrac{1}{2} .......1 \leq x \leq 2 } \atop {0..... otherwise }} \right.

∴

f_X(t)  =\left\{ \begin{array}{c} \dfrac{d}{dx}(0)=0...........

The expected value of x i

.e E(X) can now be estimated by taking the integral:

E(X) =  \int ^{\infty}_{\infty}  x f(x) \ dx

E(X) =  \int ^{1}_{- \infty}  x 0 dx + \int^2_1 \ x \dfrac{1}{2}\ dx + \int ^{\infty}_2 \ x0dx

E(X) =  \int ^{2}_{1}  x \dfrac{1}{2} dx

E(X) = \dfrac{1}{2}[\dfrac{x^2}{2}]^2_1

E(X) = \dfrac{1}{2}[\dfrac{4}{2}-\dfrac{1}{2}]

E(X) = \dfrac{1}{2} \times [\dfrac{3}{2}]

\mathbf{E(X) = \dfrac{3}{4}}

3 0
3 years ago
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