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3 years ago

8

A beaker was filled with 10 ounces of water before starting a science experiment. After the experiment, 6 ounces of water remain

ed in the beaker. How does the amount of water in the beaker before the experiment compare to the amount of water in the beaker after the experiment?
There were 16 more ounces of water before the experiment.
There were 4 fewer ounces of water before the experiment.
There were 4 fewer ounces of water after the experiment.
There were 16 more ounces of water after the experiment

2 answers:

KiRa [710]3 years ago

5 0

Answer: There were 4 fewer ounces of water after the experiment.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

Amount of water before the experiment: 10 ounces

Amount of water after the experiment: 6 ounces

10 > 6: there were fewer ounces of water after the experiment.

So, if we subtract the amount of water after the experiment (6) to the amount of water before the experiment (10), we obtain the water used for the experiment:

Mathematically speaking:

10-6 = 4

In conclusion, there were 4 fewer ounces of water after the experiment.

Allisa [31]3 years ago

3 0

There were 4 fewer ounces of water after the experiment

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A shoe store buys a pair of boots from a supplier for $45. The store's manager decides on a markup of 35%. What is the new price

Gnom [1K]

Answer:

The new price of the books = $60.75

Step-by-step explanation:

As the original price is: $45

As the markup is: 35% = 35/100 = 0.35

The Markup on the original price is:

0.35 × 45 = $15.75

Hence, the new price of the boots:

$45 + $15.75 = $60.75

Therefore, the new price of the books = $60.75

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3 years ago

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 0.95. Assuming indep

Gre4nikov [31]

Answer:

a) There is a 99.99997% probability that at least one valve opens.

b) There is a 22.62% probability that at least one valve fails to open.

Step-by-step explanation:

For each valve, there are only two possible outcomes. Either it open, or it does not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.95, n = 5$

(a) What is the probability that at least one valve opens?

Either no valves open, or at least one opens. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

So

$P(X \geq 1) = 1 - P(X = 0)$

$P(X = 0) = C_{5,0}.(0.95)^{0}.(0.05)^{5} = 0.0000003$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0000003 = 0.9999997$

There is a 99.99997% probability that at least one valve opens.

(b) What is the probability that at least one valve fails to open?

Either all valves open, or at least one does not open. The sum of the probabilities of these events is decimal 1. So

$P(X = 5) + P(X \leq 4) = 1$

We want $P(X \leq 4)$

So

$P(X \leq 4) = 1 - P(X = 5)$

$P(X = 5) = C_{5,5}.(0.95)^{5}.(0.05)^{0} = 0.7738$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.7738 = 0.2262$

There is a 22.62% probability that at least one valve fails to open.

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3 years ago

Mary has saved $24. Each week, she saves $12 more. How much money would she have 32 weeks later?

nikitadnepr [17]

Answer:

$408

Step-by-step explanation:

It’s very simple. Since gets 12 dollars every week, multiply the number of weeks and how much money she gets. Plus the amount she already had.

32 x 12 = 384

384 + 24 = 408

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2 years ago

8. Stephanie bought a pair of earrings for $3.25 and a necklace for

Marysya12 [62]

Answer:

the answer is 9.56

Step-by-step explanation:

3.25 + 5.25 = 8.50 ÷ 8 = 1.06 + 8.50 = 9.56

hope it helps :)

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3 years ago

Section 5.2 Problem 21:

Fittoniya [83]

Answer:

$y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$ (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation $am^2+bm+c=0$ where the values of $m$ are the roots:

$y''+4y'+10y=0\\\\m^2+4m+10=0\\\\m^2+4m+10-6=0-6\\\\m^2+4m+4=-6\\\\(m+2)^2=-6\\\\m+2=\pm\sqrt{6}i\\\\m=-2\pm\sqrt{6}i$

Since the values of $m$ are complex conjugate roots, then the general solution is $y(x)=e^{\alpha x}[C_1cos(\beta x)+C_2sin(\beta x)]$ where $m=\alpha\pm\beta i$ .

Thus, the general solution for our given differential equation is $y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]$ .

To account for both initial conditions, take the derivative of $y(x)$ , thus, $y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]$

Now, we can create our system of equations given our initial conditions:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(0)=e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]=3\\\\C_1=3$

$y'(x)=-2e^{-2x}[C_1cos(\sqrt{6}x+C_2sin(\sqrt{6}x)]+e^{-2x}[-C_1\sqrt{6}sin(\sqrt{6}x)+C_2\sqrt{6}cos(\sqrt{6}x)]\\\\y'(0)=-2e^{-2(0)}[C_1cos(\sqrt{6}(0))+C_2sin(\sqrt{6}(0))]+e^{-2(0)}[-C_1\sqrt{6}sin(\sqrt{6}(0))+C_2\sqrt{6}cos(\sqrt{6}(0))]=-2\\\\-2C_1+\sqrt{6}C_2=-2$

We then solve the system of equations, which becomes easy since we already know that $C_1=3$ :

$-2C_1+\sqrt{6}C_2=-2\\\\-2(3)+\sqrt{6}C_2=-2\\\\-6+\sqrt{6}C_2=-2\\\\\sqrt{6}C_2=4\\\\C_2=\frac{4}{\sqrt{6}}\\ \\C_2=\frac{4\sqrt{6}}{6}\\ \\C_2=\frac{2\sqrt{6}}{3}$

Thus, our final solution is:

$y(x)=e^{-2x}[C_1cos(\sqrt{6}x)+C_2sin(\sqrt{6}x)]\\\\y(x)=e^{-2x}[3cos(\sqrt{6}x)+\frac{2\sqrt{6}}{3}sin(\sqrt{6}x)]$

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2 years ago