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2 years ago

5

Hugo began his science experiment with 22 grams of sugar crystals. During the experiment, the change in the mass of the sugar cr

ystals was
6 grams.
What is the final mass of the sugar crystals?

1 answer:

fgiga [73]2 years ago

6 0

Answer:

I don't think there's enough information here to answer your question

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The depreciating value of a semi-truck can be modeled by y = Ao(0.83)x, where y is the remaining value of the semi, x is the tim

xenn [34]

The exponential graph is shown below.

The initial value when the time is zero is 70000. An initial value is normally shown as the point where the graph crosses the y-axis.

If the initial value was to be 50000, the curve would have crossed the y-axis at 50000

The correct answer is the first statement

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3 years ago

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What is the solution to the inequality 5.25 – b ≥ 6.5?

natita [175]

$- b \geqslant 6.5 - 5.25 \\$

$b \leqslant 1.25$

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2 years ago

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Line AC and line DB intersect at point P. Solve for angle BPQ.

taurus [48]

Answer:

Angle BPQ = 64°

Step-by-step explanation:

4x + 12 +2x = 90

6x + 12 = 90

- 12 -12

6x = 78

x = 13°

BPQ = ((4(13) + 12)°

(52 + 12)°

64°

8 0

2 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

2 years ago

A researcher wishes to estimate the number of households with two cars. How large a sample is needed in order to be 99% confiden

solmaris [256]

Answer:

Sample size n = 1382

so correct option is D) 1382

Step-by-step explanation:

given data

confidence level = 99 %

margin of error = 3%

probability = 25 %

to find out

How large a sample size needed

solution

we know here P = 25 %

so 1 - P = 1 - 0.25

1 - P = 0.75

and we know E margin of error is 0.03 so value of Z for 99%

α = 1 - 99% = 1 - 0.99

α = 0.01

and $\frac{\alpha}{2}$ = $\frac{0.01}{2}$

$\frac{\alpha}{2}$ = 0.005

so Z is here

$Z_(\frac{\alpha}{2})$ = 2.576

so

sample size will be

Sample size n = $(\frac{(Z_(\frac{\alpha}{2})}{E})^2 * P * (1-P)$

put here value

Sample size n = (\frac{2.576}{0.03})^2 * 0.25 * 0.75

Sample size n = 1382

so correct option is D) 1382

3 0

3 years ago