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3 years ago

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There are five cards in front of you, in a random order face down; 9, 10,J,Q,K, A. If you draw the 10, you lose. You have to dra

w twice in a row, what are your chances of winning?

1 answer:

salantis [7]3 years ago

5 0

Answer:

85%

Step-by-step explanation:

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Please help!<br> And explain the answers too!

RideAnS [48]

1) He worked 4 hours leaving 6 more hours for him.
rate for A alone 6 baker
rate for A & B together 4
1/a+1/b=1/x
1/6+1/b=1/4
1/b=1/4-1/6
1/b=6/24-4/24
1/b=2/24
1/b=1/12
b=12
rate for wife alone 12 hours for the remaining 6 hours
check
1/6+1/12=1/4
12/72+6/72=1/4
18/72=1/4
ok
codewa
But we want to know how long it would have taken her for the whole job not just part of it
he works twice as fast as she does so the job that takes him 10 hours alone would take her 20 hours alone.

2) jessica and cynthia work in a pet shop.
it takes jessica 6 hours to groom all the pets but cynthia needs 8 hours to groom them.
if jessica starts to groom 1 hour before cynthia joins, how long will it take them to finish?
:
let t = time it takes them to finish (C's working time)
then
t+1 = J's working time
let 1 = completed job (all pets groomed)
:
A shared work equation, each does a fraction of the job, the two fractions add up to 1
%28%28t%2B1%29%29%2F6 + t%2F8 = 1
Multiply by the least common multiple of 6 and 8, 24. cancel the denominators
4(t+1) + 3t = 24
4t + 4 + 3t = 24
4t + 3t = 24 - 4
7t = 20
t = 20/7
t = 26%2F7 hrs, which is: 2 + 6%2F7*60 = 2 hrs 51.43 min to finish the job

3) math club takes 40 minutes
science takes 50 minutes.
rate * time = quantity
quantity = 1 set up of the tables.
rate for math club is 1/40 of all the tables in 1 minute.
rate for science club is 1/50 of all the tables in 1 minute.
math club works for 10 minutes.
rate * time = quantity
1/40 * 10 = 10/40 = 1/4 of the tables are already set up by the math club.
there are 3/4 of the tables that still need to be set up.
rate * time = quantity
the rates of the math club and the science club are additive.
(1/40 + 1/50) * time = 3/4 of the tables that still need to be set up.
common denominator is 200.
(5/200 + 4/200) * time = 3/4.
9/200 * time = 3/4
time = 3/4 * 200/9 = 600 / 36 = 16 and 2/3 minutes.
total time required is 10 + 16 + 2/3 minutes = 26 + 2/3 minutes.
how to check.
10 minutes is used to do 1/4 of the tables.
16 + 2/3 minutes is used to do the remaining 3/4 of the tables.
combined rate of math and science club is 9/200 of the tables in 1 minute.
math club rate of 1/40 * 10 = 10/40 = 1/4 of the tables.
combine rate of 9/200 * (16 + 2/3) = 9/200 * 50/3 = 450/600 = 9/12 = 3/4 of the tables.
1/4 + 3/4 = 1 = all of the tables.

6 0

3 years ago

What is the VOLUME and SURFACE AREA of a cube with 14 cm in length?

vfiekz [6]

Volume of the cube = a³ = (14cm)³= 2744 cm³

To find surface are, we need to find at first area of the face
A(face)=14² cm².
Cube has 6 congruent squares, so
Surface area of the cube = 6*14² cm²= 1176 cm²

3 0

3 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago

Edge- a line segment where two __ meet

UNO [17]

Hey there!

$Answer:\boxed{\text{verticles}}$

Edge - a line segment where two <u>verticals</u> meet.

Hope this helps!

$\text{-TestedHyperr}$

3 0

3 years ago

Will someone please go answer my other question??!!‍♀️

abruzzese [7]

The answer to your question is YES!

4 0

3 years ago