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2 years ago

When is the best time to use square roots to solve a quadratic? Choose the one BEST

Mathematics

1 answer:

blondinia [14]2 years ago

8 0

Answer: D, When the constants are perfect squares.

Step-by-step explanation:

the “best” method whenever the quadratic equation only contains x2 terms. That implies no presence of any x term being raised to the first power somewhere in the equation.

Hopefully this helps!

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In a certain community, 30% of the families own a dog, and 20% of the families that own a dog also own a cat. It is also known t

Nesterboy [21]

Answer:

a) 0.06

b) 0.778

Step-by-step explanation:

Let's suppose a community of 100 families just to facilitate the calculation.

30% of the families own a dog

Dog = 30% of 100 = 30

20% of the families that own a dog also own a cat = 20% of 30 = 6

27% of all the families own a cat = 27% of 100 = 27

So, 6 families own a dog and a cat.

As 30 families own a dog, [30 - 6 =] 24 families own only dogs

As 27 families own a cat, [27 - 6 = ] 21 families own only cats

See picture attached.

a) What is the probability that a randomly selected family owns both a dog and a cat?

P(dog and cat) = dog ∩ cat/total = 6/100 = 0.06

b) What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?

So, only cat/total cat

P (not dog|cat) = 21/27 = 0.778

3 0

3 years ago

Which expression is equivalent to (64y^100)^1/2

marin [14]

Answer:

8y^50

Step-by-step explanation:

7 0

3 years ago

Find out the number of combinations and the number of permutations for 8 objects taken 6 at a time. Express your answer in exact

umka2103 [35]

Solution:

The permutation formula is expressed as

$\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\ \end{gathered}$

The combination formula is expressed as

$\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\ \\ \end{gathered}$

where

$\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}$

Given that 6 objects are taken at a time from 8, this implies that

$\begin{gathered} n=8 \\ r=6 \end{gathered}$

Thus,

Number of permuations:

$\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}$

Number of combinations:

$\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}$

Hence, there are 28 combinations and 20160 permutations.

7 0

1 year ago

can someone help me out i dont know what the answer is an i suck at this ive already done it once and i dont know how to do this

hoa [83]

Answer:

-3

Step-by-step explanation:

When you have an equation in the form y=mx+b, then the number at the end is the y-intercept.

7 0

2 years ago

Read 2 more answers

Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.

Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes = (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0

3 years ago