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3 years ago

14

A collection of coins contains a total of sixteen dimes and quarters. It is worth $2.20 in all. How many dimes are in the collec

tion?

1 answer:

grin007 [14]3 years ago

5 0

9514 1404 393

Answer:

12 dimes

Step-by-step explanation:

Let q represent the number of quarters. Then the number of dimes is 16-q and the total value is ...

0.25q +0.10(16 -q) = 2.20

0.15q +1.60 = 2.20 . . . . . . . simplify

0.15q = 0.60 . . . . . . . . subtract 1.60

q = 4 . . . . . . . . . . . divide by 0.15

16-q = 12

There are 12 dimes in the collection.

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A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

2 years ago

What is the domain of g?

lawyer [7]

Answer:

A choice.

Step-by-step explanation:

Domain starts at x = - 7 and ends at x = 4. The domain from the graph is also continuous. Therefore, we can rule out C and D.

B is not correct as domain starts from x = -7 and not x = - 4.

8 0

2 years ago

Determine the point estimate of the population proportion and the margin of error for the following confidence interval.Lower bo

Kamila [148]

Answer: The point estimate of the population proportion is . 485.

The margin of error is 0.273.

Step-by-step explanation:

Confidence interval for population proportion(p):

sample proportion ± Margin of error

Given: Lower bound of confidence interval = 0.212

Upper bound = 0.758

⇒sample proportion - Margin of error=0.212 (i)

sample proportion + Margin of error= 0.758 (ii)

Adding (i) and (ii) , we get

2(sample proportion) =0.970

⇒ sample proportion = 0.970÷2= 0.485

Since sample proportion is the point estimate of the population proportion.

So, the point estimate of the population proportion= 0.485

Now put sample proportion =0.485 in (ii), we get

0.485+ Margin of error= 0.758

⇒ Margin of error= 0.758 - 0.485 =0.273

i.e. The margin of error is 0.273.

5 0

3 years ago

P, Q & R form a right-angled triangle.

Naya [18.7K]

Answer:

m<RPQ = 22°

Step-by-step explanation:

Given:

m<SRQ = 90°

PS = PQ

m<SQR = 46°

Required:

m<RPQ

Solution:

m<SQR + m<SRQ + m<RSQ = 180°

Substitute

46° + 90° + m<RSQ = 180°

m<RSQ = 180° - 136°

m<RSQ = 44°

Find m<PSQ:

m<PSQ = 180° - m<RSQ (Angles on a straight line

m<PSQ = 180° - 44° (Substitution)

m<PSQ = 136°

Find m<RPQ:

∆QSP is an isosceles triangle with two equal base angles. Therefore:

m<RPQ = ½(180° - 136°)

m<RPQ = 22°

4 0

3 years ago

Help me with my math please

sergij07 [2.7K]

Step-by-step explanation:

1 _21:00_ , _22:10_, 1:10

2. 1:20

3. 18:00

4. 21:35

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2 years ago

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