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3 years ago

11

Jody is buying a scrapbook and sheets of designer paper. She has $40 and needs at least $18.25 to buy the scrapbook. Each sheet

of paper costs $0.34. How many sheets of paper can she buy? Show your work.

1 answer:

Harlamova29_29 [7]3 years ago

7 0

Jody can buy 63 sheets of paper. $40- 18.25= 21.75. 21.75 divided by .34= 63. To double check you would add 18.25+ .34(63) = 39.67

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Q2) What are the rules for probability distributions ? I

olga2289 [7]

Answer:

Probability Distributions

A listing of all the values the random variable can assume with their corresponding probabilities make a probability distribution.

A note about random variables. A random variable does not mean that the values can be anything (a random number). Random variables have a well defined set of outcomes and well defined probabilities for the occurrence of each outcome. The random refers to the fact that the outcomes happen by chance -- that is, you don't know which outcome will occur next.

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3 years ago

Evaluate the following algebraic expression, if (x = 5, y = 3)<br> x² +9y - 5

gladu [14]

Answer: 47

Step-by-step explanation:

simply substitute the constants with 5 and 3.

(5)^2 + 9(3) - 5 = 47

3 0

3 years ago

Read 2 more answers

Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation? Solve x2 = 12x – 15 by completing the s

Svetlanka [38]

Answer: $6+\sqrt{21}\ \text{and}\ 6-\sqrt{21}$

Step-by-step explanation:

Given

Quadratic equation is

$x^2-12x+15=0$

Solving by completing the square method

$\Rightarrow x^2-2\cdot 6\cdot x+15=0\\\text{add and subtract 36}\\\Rightarrow x^2-2\cdot 6\cdot x+15+36-36=0\\\Rightarrow x^2-2\cdot 6\cdot x+36+(-36+15)=0\\\text{using}\ (a-b)^2=a^2+b^2-2ab\\\Rightarrow (x-6)^2-21=0\\\Rightarrow (x-6)^2=21\\\Rightarrow x-6=\pm \sqrt{21}\\\Rightarrow x=6\pm \sqrt{21}$

The solution set of the equation is $6+\sqrt{21}\ \text{and}\ 6-\sqrt{21}$

6 0

3 years ago

Read 2 more answers

84/12 = 28/x 1. x=4 2. X=28

yan [13]

Answer:

x = 4

Step-by-step explanation:

We can use cross products to solve

84/12 = 28/x

84 * x = 12 *28

Divide each side by 84

84x/84 = 12*28/84

x =4

5 0

3 years ago

A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min

Dmitry [639]

At any time $t$ (min), the volume of solution in the tank is

$500\,\mathrm{gal}+\left(5\dfrac{\rm gal}{\rm min}-5\dfrac{\rm gal}{\rm min}\right)t=500\,\mathrm{gal}$

If $A(t)$ is the amount of salt in the tank at any time $t$ , then the solution has a concentration of $\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}$ .

The net rate of change of the amount of salt in the solution, $A'(t)$ , is the difference between the amount flowing in and the amount getting pumped out:

$A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)$

Dropping the units and simplifying, we get the linear ODE

$A'=10+5\sin\dfrac t4-\dfrac A{10}$

$10A'+A=100+50\sin\dfrac t4$

Multiplying both sides by $e^{10t}$ allows us to identify the left side as a derivative of a product:

$10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}$

$\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}$

$e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt$

Integrate and divide both sides by $e^{10t}$ to get

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}$

The tanks starts off with 30 lb of salt, so $A(0)=30$ and we can solve for $C$ to get a particular solution of

$A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}$

6 0

4 years ago