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3 years ago

14

Congratulations! You won a free grill in a sweepstakes! You know that you can fit a rectangular grill in your truck that has a w

idth of 2 1/2 feet. The length of the grill can be up to three times the width. What is the largest possible perimeter your new grill can have to fit in your taco truck? Show your work.

1 answer:

Basile [38]3 years ago

8 0

Answer: 20 feet perimeter is the largest allowed

Step-by-step explanation: width=2.5 length=no more than 3x2.5

so find longest possible length allowed...

2.5x3=7.5

calculate largest perimeter allowed...

2(7.5+2.5)=20

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6. 200 pages to 600 pages<br>percentage of change

adell [148]

Answer:

What

Step-by-step explanation:

7 0

3 years ago

Imelda jogs 2 3/4 miles each weekday, Monday through Friday. On Saturday, she logs 2.5 times the daily weekday distance. On Sund

Colt1911 [192]

Answer: 22.6875 miles

Step-by-step explanation:

Monday = 2 3/4 miles = 2.75 miles

Tuesday = 2.75 miles

Wednesday = 2.75 miles

Thursday = 2.75 miles

Friday = 2.75 miles

Saturday = 2.5 × 2.75 = 6.875 miles

Sunday = 0.75 × 2.75 = 2.0625 miles

Total = 2.75 miles + 2.75 miles + 2.75 miles + 2.75 miles + 2.75 miles + 6.875 miles + 2.0625 miles

Total = 22.6875 miles

5 0

3 years ago

Brianna is going to a carnival that has games and rides. Each game costs $1.25 and each ride costs $3.75. Brianna spent $18.75 a

maxonik [38]

Answer:

6 games and 3 rides

Step-by-step explanation:

Let the number of games Brianna played be x,

and the number of rides she went on be y.

Total cost for games= 1.25x

Total cost for rides= 3.75y

Since number of games is twice the number of rides,

x= 2y -----(1)

Total costs= cost of games +cost of rides

$18.75= 1.25x +3.75y$

Multiply the whole equation by 4 to remove the decimals:

$75= 5x +15y$

Simplify by dividing the whole equation by 5:

$15 = x + 3y$

Label the equation:

x +3y= 15 -----(2)

Although we can solve these 2 equations by substitution, since the question requires us to graphically solve, we have to graph 2 linear lines.

I will choose 3 points to plot on the graph for each equation:

x= 2y -----(1)

$\begin{tabular}{|c|c|c|c|} \cline{1-4}x & 2(1) = 2&2(2) = 4&2(3) = 6\\ \cline{1-4}y & 1 &2&3\\ \cline{1-4}\end{tabular}$

x +3y= 15 -----(2)

x= -3y +15

$\begin{tabular}{|c|c|c|c|} \cline{1-4}x & -3(1)+15= 12& -3(2)+15= 9& -3(3)+15= 6\\ \cline{1-4}y & 1 &2&3\\ \cline{1-4}\end{tabular}$

Let's plot these points on a graph paper. Then, join them with a straight line for each straight line graph. Please see the attached picture for the graph.

From (1): y= ½x

From (2): 3y= 15 -x

y= 5 -⅓x

From the graph, the solution of the equation is (6,3). The solution is the point on the graph in which the 2 lines intersect.

x- coordinate: 6

y- coordinate: 3

Thus, Brianna played 6 games and went on 3 rides.

8 0

4 years ago

Read 2 more answers

Helppoo 10 points! I’ll mark brainliest!

77julia77 [94]

Answer Yes, because you take the answer from 100+45 and subtract 3 to get to 97 and then you add 29 to make the 47 a 74.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

5 0

4 years ago