Find the quotient. 36s^3t-26st ----------------- -2st Enter the correct answer.

What is the range for the following set of numbers?57, -5, 11, 39, 56, 82, -2, 11, 64, 18, 37, 15, 68

antiseptic1488 [7]

<h3> The range= the highest number - the lowest number</h3>

so

82-(-2)

=84

then ur answer is 84

7 0

3 years ago

If DEF and MNO are two triangles such that DE=MN and EF=NO, which of the following would be sufficient to prove that triangles a

Delvig [45]

Answer:

D. ∠E ≅ ∠N

Step-by-step explanation:

The pair of sides meet at vertex E in ∆DEF and at vertex N in ∆MNO. Since the sides that make up angles E and N are shown congruent, it is sufficient to show ...

∠E ≅ ∠N

Then the SAS congruence postulate can be claimed.

__

Additional comment

The alternative is to show DF ≅ MO. That would allow you to claim SSS congruence. That is not an answer choice.

3 0

3 years ago

A system has one each of two different types of components in joint operation. Let X and Y denote the random lengths of the live

RSB [31]

Answer:

The value of E(Y/X) is 2.

Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

$f(x,y)=\left \{ {{\dfrac{x}{8}e^{-\dfrac{x+y}{2}} \,\,\,\,0 \leq x,0\leq y \atop {0}} \right.$

So the marginal function for X is given as

$f_x=\int\limits^{\infty}_0 {f(x,y)} \, dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x+y}{2} }dy\\f_x=\int\limits^{\infty}_0 \dfrac{x}{8}e^{-\frac{x}{2}}e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{8}e^{-\frac{x}{2}}\int\limits^{\infty}_0e^{-\frac{y}{2} }dy\\f_x= \dfrac{x}{4}e^{-\frac{x}{2}}$

Now

$f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }$

Now the value of E(Y/X) is given as

$E(Y/X)=\int\limits^{\infty}_0 {yf_{Y/X}} \, dy \\E(Y/X)=\int\limits^{\infty}_0 {y\dfrac{1}{2}e^{-\frac{y}{2} } }\, dy\\E(Y/X)=\dfrac{1}{2}\dfrac{\sqrt{2}}{(\dfrac{1}{2})^2}=2$

So the value of E(Y/X) is 2.

3 0

3 years ago

Beaker A contains 1 liter which is 15 percent oil and the rest is vinegar, thoroughly mixed up. Beaker B contains 2 liters which

Tanzania [10]

Answer:

Add 1.25 litres of oil.

Step-by-step explanation:

Contents in beaker A implies;

15% of oil in 1 litre = 0.15 litres of oil

So that, there are 0.15 litres of oil and 0.85 litres of vinegar in beaker A.

Contents in beaker B implies:

55% of oil in 2 litres = 1.1 litres of oil

So that, thee are 1.1 litres of oil and 0.9 litres of vinegar in beaker B.

Half of contents of B poured into A implies that beaker A now contains:

0.15 litres + 0.55 litres = 0.7 litres of oil

0.85 litres + 0.45 litres = 1.3 litres of vinegar

The percentage of oil in A now = $\frac{0.7}{2}$ x 100%

= 35%

To increase the percentage of oil to 60%, then:

0.7 litres + 1.25 litres = 1.95 litres of oil

But the new total litres of the content in beaker A = 3.25 litres

So that;

$\frac{1.95}{3.25}$ x 100% = 60%

Therefore, to produce a mixture which is 60% oil, add 1.25 litres of oil.

7 0

3 years ago

Which of the following word problems can be solved using the equation 10 x - 15=60

Alja [10]

Answer:

Kit buys

Step-by-step explanation:

im right

8 0

3 years ago