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mars1129 [50]
3 years ago
11

1. A total of 498 tickets were sold for the school play. They were either adult tickets or student tickets. There were

Mathematics
2 answers:
bonufazy [111]3 years ago
7 0
Out of 498 tickets, 446 were adults
498 - 52 = 446

QveST [7]3 years ago
3 0
What you do to solve this is:
1- 498 - 52 = 446.
2- 446 adult tickets were sold.

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FIVE STARS AND BRAINLIEST FOR CORRECT ANSWER
Charra [1.4K]

For the given vectors \vec{a}=a_{1}\hat{i}+a_{2}\hat{j} and \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}

The dot product of vectors a and b is defined as = \vec{a}.\vec{b}=(a_{1}\times b_{1}+a_{2} \times b_{2})

So, \vec{a}.\vec{b} = (4\hat{i}+3\hat{j}). (-4\hat{i}+4\hat{j})

\vec{a}.\vec{b}=(4\times(-4)+3 \times4)

= -16+12

= -4

8 0
3 years ago
Gage's math teacher entered the seventh-grade student in a math competition. There was an enrollment fee of $30 and also $11 cha
Amanda [17]

Answer:

110 test

Step-by-step explanation:

Let

x ----> the number  of packet of 10 test

y ----> the total cost

we know that

The number of packet of 10 test purchase multiplied by $11 plus the enrollment fee of $30 must be equal to $151

so

The linear equation that represent this scenario is

y=11x+30

we have

y=\$151

substitute

151=11x+30

Solve for x

subtract 30 both sides

11x=151-30

11x=121

Divide by 11 both sides

x=11

The number of packets purchase was 11

To find out the number of test, multiply the number of packets by 10

11(10)=110\ test

4 0
3 years ago
The ideal gas law relates the pressure P , volume V , and temperature T of a fixed number of molecules of a gas in a container v
pochemuha

Answer:

K would be 375.

Step-by-step explanation:

You take the number you already have and divide it by the new number its giving. (0.06)

8 0
3 years ago
Select the correct answer.<br> What is the domain of the function shown on the graph?
maw [93]

Answer:

B. negative infinity < x < positive infinity

Step-by-step explanation:

When a question asks for the domain of a function, it is asking for all possible x-values, so we can rule out choice A since there are clearly x-values greater than 1. This graph shows arrows pointing left and downward, indicating that the graph continues past what we can see. This tells us that even though the x-values look like they end at -10 and 2, they continue further, so we can also rule out option C and option D. After these eliminations, we are left with choice B, which is a way of writing "All real numbers," and is the domain for <u>all</u> exponential functions, including this one.

lmk if im incorrect about anything, hope this helps :)

5 0
3 years ago
Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.
Cloud [144]

If F_Y(y) is the cumulative distribution function for Y, then

F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)

Then the probability density function for Y is f_Y(y)={F_Y}'(y):

f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

The nth moment of Y is

E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy

Let u=\ln y, so that \mathrm du=\frac{\mathrm dy}y and y^n=e^{nu}:

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du

Complete the square in the exponent:

nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2

E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du

But \frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2} is exactly the PDF of a normal distribution with mean n and variance 1; in other words, the 0th moment of a random variable U\sim N(n,1):

E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1

so we end up with

E[Y^n]=e^{\frac12n^2}

3 0
3 years ago
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