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3 years ago

Pam and her sister have a pattern of saving money. Pam saved 40 dollars and her brother saved 100 dollars.

Mathematics

2 answers:

Keith_Richards [23]3 years ago

8 0

Answer:

true

Step-by-step explanation:

Pam saved 40

her bro saves 100

let us take this as ration 40 is to 100 which can also be written in fractional for like 40/100 now we simplify it how by dividing both by 20 so (40/20)/(100/20) which will give 2/5 which can be written as 2 is to 5 so this is true

Marysya12 [62]3 years ago

4 0

Answer:

Oct 25, 2016 — Pam and her brother both open savings accounts. each begin with a balance of $0. For every $2 is that Pam saves in her account her brother ...

Step-by-step explanation:

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Veronika [31]

Answer:

the answer is c)SAS

Step-by-step explanation:

as the sides are already given and and both angles will be equal as vertically opposite angles (VOA)are always equal.

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4 years ago

What is the exact value of cos(11pi/21)cos(pi/7)-sin(11pi/21)sin(pi/7)?

Anika [276]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

According to the graph, what is the value of the constant in the equation

wolverine [178]

The question is missing the graph. So, it is attached below.

Answer:

(D) 3.2

Step-by-step explanation:

Given:

A graph of height versus width.

The equation given is:

Height = constant × Width

Rewriting in terms of 'constant'. This gives,

$\textrm{Constant}=\frac{Height}{Width}$ ------------- (1)

The width is plotted on the X-axis and the corresponding height is plotted on the Y-axis.

The four points plotted on the line are:

$(x, y) = (0.5, 1.6), (1, 3.2), (2, 6.4)\ and\ (3, 9.6)$ .

Now, any point will satisfy equation (1).

Consider the point (0.5, 1.6). So, height = 1.6 and width = 0.5. Therefore,

$\textrm{Constant}=\frac{Height}{Width}\\\\\textrm{Constant}=\frac{1.6}{0.5}=3.2$

Also, we observe that for all the remaining points,

$\dfrac{3.2}{1}=\dfrac{6.4}{2}=\dfrac{9.6}{3}=3.2$ .

Hence, the value of the constant is 3.2.

Option (D) is correct.

4 0

3 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago

This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my w

Sedaia [141]

Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)
2pi*h^2*dh = 25*dt
int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides
(2/3)pi*h^3 = 25t + C
2pi*h^3 = 3(25t + C)
h^3 = (3(25t + C))/(2pi)
h^3 = (75t + 3C)/(2pi)
h^3 = (75t + C)/(2pi)
h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0
0 = [ (75(0) + C)/(2pi) ]^(1/3)
0 = [ (0 + C)/(2pi) ]^(1/3)
0 = [ (C)/(2pi) ]^(1/3)
0^3 = (C)/(2pi)
0 = C/(2pi)
C/(2pi) = 0
C = 0*2pi
C = 0

Therefore the h(t) function is...
h(t) = [ (75t + C)/(2pi) ]^(1/3)
h(t) = [ (75t + 0)/(2pi) ]^(1/3)
h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct.
Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
75t = 1024pi
t = 1024pi/75
At this time value, the height of the water is 8 feet

Set up the radius r(t) function
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)

8 0

3 years ago