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3 years ago

I just need help in this last question plsssss

Mathematics

1 answer:

finlep [7]3 years ago

8 0

Answer:

B) 0.8

Step-by-step explanation:

If events are said to be independent, then just multiply the probability of the first event by the second if they are supposed to be happening together.

P (A) * P (B) = 0.8 x 1 = 0.8

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Josh is hiking Glacier National Park. He has now hiked a total of 17 \text{ km}17 km17, start text, space, k, m, end text and is

EastWind [94]

Answer:

The equation to determine the total length in kilometers is $\frac{h}{2} =(17+2)$

The total length in kilometers of Josh’s hike is 38 km.

Step-by-step explanation:

Given:

Let the total length in kilometers of Josh’s hike be h.

Now Given that He has now hiked a total of 17 km and is 2 km short of being 1/2 of the way done with his hike.

It means that to reach half of the length of total length Josh needs 2 more km to add in his hiking which is done which is of 17 km.

Framing the above sentence in equation form we get;

$\frac{h}{2}=(17+2)$

Hence, The equation to determine the total length in kilometers is $\frac{h}{2} =(17+2)$

Now Solving the above equation we get;

First we will multiply 2 on both side using Multiplication property we get;

$2\times \frac{h}{2}= 2\times(17+2)\\\\h =2\times 19 =38 \ km$

Hence, The total length in kilometers of Josh’s hike is 38 km.

4 0

3 years ago

Read 2 more answers

Omg, I need help! A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6

klemol [59]

Answer:

Average size of the lots = ⅓ acre

Step-by-step explanation:

The question incomplete without specifying what we are to determine.

Question:A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X's. Organize the information in a line plot. What is the average size of the lots? _________ acre

Help anyone?

Solution:

We are asked to organize the information in a line plot. See attachment for the line plot.

Given: 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X'sIn no particular order, the sizes of the lots are:1/6, 1/6, 1/6, 1/6, 1/6, 1/6, 1/3, 1/3, 1/3, 1/2, 1/2, 1/2, 1/2, 1/2 and 1/2 acre.

Let's count the number of lot for each size given.

For 1/6: there are 6 X's on the line plot of 1/6 number of lot for 1/6

= the lot × number of times it occurs

= (1/6) × 6 = 6/6 = 1 acre

For 1/3: there are 3 X's on the line plot of 1/3 number of lot for 1/3 = the lot × number of times it occurs

= (1/3) × 3 = 3/3 = 1 acre

For 1/2: there are 6 X's on the line plot of 1/2 number of lot for 1/2

= the lot × number of times it occurs

= (1/2) × 6 = 6/2= 3 acres

To find average size of the lots, we would sum all lot for each given size then divide by the total number of lots given.Sum of all lot for each given size = 1+1+3

Sum of all lot for each given size = 5

The total number of lots given = 15

Average size of the lots = 5/15 = 1/3

Average size of the lots = ⅓ acre

7 0

3 years ago

X over 2 +2= 15 two step equations

Fed [463]

Answer:

$\boxed{\boxed{\sf x=84}}$

Step-by-step explanation:

$\sf \cfrac{x}{6} +2=16$

Multiply both sides of the equation by 6:

$\sf x+12=96$

Subtract 12 from both sides:

$\sf x=96-12$

Subtract 12 from 96:

$\sf x=84$

_________________________________

3 0

2 years ago

Read 2 more answers

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: