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Umnica [9.8K]
2 years ago
3

See clear window cleaner is 40% vinegar and spot free window cleaner is 90% vinegar. How much of each should be used to make a 2

50 oz of a cleaner that is 60% vinegar?
Mathematics
1 answer:
nirvana33 [79]2 years ago
5 0
Okay so I would say to do
x = clear window cleaner and y = spot free window cleaner

x+y = 250
.4x+.9y=(.6)250 so .4x+.9y=150

We are going to use the substitution technique here so x=250-y from the top equation. Substitute that into the bottom

.4(250-y)+.9y=150 so now 100-.4y+.9y=150. 0.9-.4=.5 and 150-100 = 50
therefore
.5y=50
Divide by .5 and get 100 = y, substitute that back into x=250-y and get
x=250-100, so x =150.

Final answers: 

150 ounces of clear window cleaner and 100 ounces of spot free window cleaner
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Three dimensional figures
aksik [14]

Answer:

3d figures - "a three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth." -Google

Some Examples are...

- cones

-cylinders

-cubes

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Hope this helps!

4 0
2 years ago
the length of the highway is 1500 miles. If 1 inch represents 150 miles, what is the length of the highway on the map?
NeX [460]

Answer:

10 inches.

Step-by-step explanation:

If 1 inch represents 150 miles, and there are 1500 miles covered by the map, then the length of the highway on the map is 10 inches.

4 0
2 years ago
Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancelations on different days are in
andrezito [222]
<h3>(a)Probability of cancelling at least 4 classes  is 0.0055.</h3><h3>(b)Probability of 10th class is third class that gets cancelled  = 0.0031</h3>

Step-by-step explanation:

Here, the question is<u> incomplete</u>.

Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancellations on different days are independent.

(a) There are 15 classes left this semester. Compute the probability that at least 4 of them get canceled.

(b) Compute the probability that the tenth class this semester is the third class that gets canceled.

Now, here:

The probability of cancelling each class  = 0.05

Now, probability of cancelling at least 4 classes

= 1 - P(Cancelling  at max 3 classes)

=  1 - P(0 ≤ x  ≤  3)   = 1 - Binomial (15,0.05,3)

= 0.0055

Hence, probability of cancelling at least 4 classes  is 0.0055.

(b) As given, 10th class is third class that gets cancelled.

So, the first and second classes that get cancelled in between 1 - 9.

P(2 class cancelled in 1st 9) = Binomial (15,0.05,3) = 0.0629

Also, as given  P(10th class cancelled) = 0.05

⇒ P(10th class is third class that gets cancelled ) = 0.0629 x 0.05 = 0.0031

5 0
2 years ago
What is 2 divided 1 6/7 ? Please answer
Aliun [14]
If you would like to divide 2 by 1 6/7, you can do this using the following steps:

1 6/7 = 13/7

2 / 1 6/7 = 2 / 13/7 = 2/1 / 13/7 = 2/1 * 7/13 = 14/13 = 1 1/13

The correct result would be 1 1/13.
6 0
2 years ago
A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon
posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x

(concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x

(concentration of substance in liquid exiting)

Let y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)}  \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)}  \:\frac{pound}{s}

V(t) is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

V(t)=100 + 2t

We can then write the initial value problem:

\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

\frac{dy}{dx}+P(x)y =Q(x)

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, \mu \left( x \right), using \mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}

\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}

Step 2: Multiply everything in the differential equation by \mu \left( x \right) and verify that the left side becomes the product rule \left( {\mu \left( t \right)y\left( t \right)} \right)' and write it as such.

\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}

Step 3: Integrate both sides.

\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C

Step 4: Find the value of the constant and solve for the solution y(t).

50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000

y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}

Now, the tank is full of brine when:

V(t) = 400\\100+2t=400\\t=150

The amount of salt in the tank when it is full of brine is

y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75

6 0
2 years ago
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