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sergij07 [2.7K]
2 years ago
11

Paulina is remodeling her bathroom. the tile she has chosen are squares and trapezoids. the side length of each square inthe til

e is x centimeters. the height and the length of one of the bases of each trapezoid is x centimeters. the other length is 2x centimeters. Write a simplified equation to solve for x in terms of At, the area of the tile. if necessary, use rational coefficients instead of root symbols

Mathematics
1 answer:
NemiM [27]2 years ago
8 0
Check the picture below.

is not very specific above, but sounds like it's asking for an equation for the trapezoid only, mind you, there are square tiles too.

but let's do the trapezoid area then, 

\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
-------------------------------\\\\

\bf A=\cfrac{h(a+b)}{2}\quad 
\begin{cases}
A=At\\
a=x\\
h=x\\
b=2x
\end{cases}\implies At=\cfrac{x(x+2x)}{2}
\\\\\\
At=\cfrac{x(3x)}{2}\implies At=\cfrac{3x^2}{2}\impliedby \textit{now, solving for \underline{x}}
\\\\\\
2At=3x^2\implies \cfrac{2At}{3}=x^2\implies \sqrt{\cfrac{2At}{3}}=x\implies \left( \frac{2At}{3} \right)^{\frac{1}{2}}=x

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
2 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
2 years ago
The figure shows the design of a greenhouse with a rectangular floor. The front and back sides of the greenhouse are semicircles
Ksju [112]

9514 1404 393

Answer:

  96 cubic feet

Step-by-step explanation:

A volume that is 16 ft by 18 ft by 1/3 ft will be ...

  V = LWH

  V = (16 ft)(18 ft)(1/3 ft) = 96 ft³

96 cubic feet of concrete are needed.

7 0
2 years ago
You invested a total of $12,000 at 4.5 percent and 5 percent simple interest. During one year, the two accounts earned $570. How
Luda [366]

Answer: You invested $6000 in both accounts.

Step-by-step explanation:

Let x represent the amount invested in the account earning 4.5% interest.

Let y represent the amount invested in the account earning 5% interest.

You invested a total of $12,000 at 4.5 percent and 5 percent simple interest. This means that

x + y = 12000

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Considering the account earning 4.5%

I = (x × 4.5 × 1)/100 = 0.045x

Considering the account earning 5%

I = (y × 5 × 1)/100 = 0.05y

During one year, the two accounts earned $570. . This means that

0.045x + 0.05y = 570 - - - - - - - - - - 1

Substituting x = 12000 - y into equation 1, it becomes

0.045(12000 - y) + 0.05y = 570

540 - 0.045y + 0.05y = 570

- 0.045y + 0.05y = 570 - 540

0.005y = 30

y = 30/0.005 = 6000

x = 12000 - y = 12000 - 6000

x = $6000

6 0
2 years ago
If a certain number is increased by 5 one half of the result is three-fifths of the excess of 61 over the number find the number
PSYCHO15rus [73]

Answer:

= 391/11

Let say number = N

a certain number is increased by 5

= N + 5

,one-half of the result

= (N + 5)/2

Three -fifths of the excess of 61 over the number.

= (3/5)(61 - N)

Equating both

(N + 5)/2  = (3/5)(61 - N)

multiplying by 10 both sides

=> 5(N + 5) = 6(61 - N)

=> 5N + 25 = 366 - 6N

=> 11N = 391

=> N = 391/11

Step-by-step explanation:

8 0
2 years ago
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