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tamaranim1 [39]
1 year ago
9

A Shampoo Problem: A bottle contains 25.4 fluid ounces of shampoo. Katie uses 0.25 of the bottle. How much shampoo is left? a. I

s the shampoo problem a word problem for 0.25 * 25.4, is it a word problem for 25.4 − 0.25, or is it not a word problem for either of these? Explain. b. Write a new shampoo word problem for 0.25 * 25.4 and write a new shampoo word problem for 25.4 − 0.25. Make clear which is which.
Mathematics
1 answer:
kotykmax [81]1 year ago
7 0

Answer:

a) There is no a word problem for both expressions (m = 0.25\cdot 25.4 and m = 25.4-0.25), b) A bottle contains 25.4 fluid ounces of shampoo. Katie uses 0.75 of the bottle. How much shampoo is left? A bottle contains 25.4 fluid ounces of shampoo. Katie uses 0.25 fluid ounces of the bottle. How much shampoo is left?

Step-by-step explanation:

a) The shampoo problem is a word problem for:

(Final content) = (Initial content) - (Used content)

Then,

m = 25.4 - 0.25\cdot 25.4

Or:

m = 0.75\cdot 25.4

Hence, there is no a word problem for both expressions (m = 0.25\cdot 25.4 and m = 25.4-0.25).

b) The word problem for m = 0.25\cdot 25.4 is:

A bottle contains 25.4 fluid ounces of shampoo. Katie uses 0.75 of the bottle. How much shampoo is left?

The word problem for m = 25.4-0.25 is:

A bottle contains 25.4 fluid ounces of shampoo. Katie uses 0.25 fluid ounces of the bottle. How much shampoo is left?

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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
High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capa
zalisa [80]

Answer:

The students have to score 0.74 standard deviations above the mean to be publicly recognized.

Step-by-step explanation:

A random variable <em>X</em> is said to have a normal distribution with parameters <em>µ</em>    (mean) and <em>σ</em>² (variance).

If X \sim N (\mu, \sigma^{2}), then z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (<em>Z</em>) = 0 and Var (<em>Z</em>) = 1. That is, Z \sim N (0, 1).

The distribution of these <em>z</em>-scores is known as the standard normal distribution.

The <em>z</em>-score is a standardized form of the raw score, <em>X</em>. It is a numerical measurement of the relationship between a value (<em>X</em>) and the mean (<em>µ</em>) in terms of the standard deviation (<em>σ</em>). A <em>z</em>-score of -1 implies that the data value is 1 standard deviation below the mean. And a <em>z</em>-score of 1 implies that the data value is 1 standard deviation above the mean.

Let <em>X</em>  be defined as the scores of students at the National Financial Capability Challenge Exam.

It is provided that the students who score in the top 23% are recognized publicly for their achievement by the Department of the Treasury.

That is, P (X > x) = 0.23.

⇒ P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma})=0.23

⇒

   P(Z>z)=0.23\\1-P(Z

The value of <em>z</em> for this probability value is:

<em>z</em> = 0.74.

*Use a <em>z</em>-table.

Thus, the students have to score 0.74 standard deviations above the mean to be publicly recognized.

8 0
2 years ago
Which expression is equivalent to square root 64a^6
zhenek [66]
I believe the answer is A
4 0
1 year ago
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Points E, F, and D are located on circle C. Circle C is shown. Line segments E C and D F are radii. Lines are drawn from points
elena55 [62]

Answer:

Option (1). 34°

Step-by-step explanation:

From the figure attached, CE and CD are the radii of the circle C.

Central angle CED formed by the intercepted arc DE = 68°

Since measure of an arc = central angle formed by the intercepted arc

Therefore, m∠CED = 68°

Since m∠EFD = \frac{1}{2}(m\angle ECD) [Central angle of an intercepted arc measure  the double of the inscribed angle by the same arc]

Therefore, m∠EFD = \frac{1}{2}(68)

                               = 34°

Therefore, Option (1) 34° will be the answer.

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1 year ago
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Given two terms in a geometric sequence, find the 8th term and the recursive formula.
Shtirlitz [24]

Answer:

Step-by-step explanation:

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We can find the explicit formula for this using the info we have about the 5th term and the 3rd term. The explicit formula looks like this, in general terms:

a_n=a_1*r^{n-1 where a1 is the first term and r is the common ratio. We can solve for both, one at a time. First, using the fact that the 5th term is 3888:

a_5=a_1*r^{5-1 and subbing in and simplifying:

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Now the other equation we need for this system, using the fact that the third term is 108:

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a_1=\frac{3888}{6^4} so

a1 = 3. Now we have the a and r for the explicit formula:

a_n=3*6^{n-1 and we simply put in an  for n:

a_8=3*6^{8-1 and

a_8=3*6^7 which gives us the 8th term as

a8 = 839808

The recursive formula is

a_n=a_{n-1}*r and it works, also, but again, you have to have all the terms that come before the one you want or this formula is useless. Explicit is always better, in my opinion.

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2 years ago
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