X +10=14 <--how does one solve this?

You roll a number cube one time.<br> What is the probability of not rolling a 8 or a 10?

liq [111]

Answer: 80% or 4/5

Step-by-step explanation:

so I don’t know how many numbers there are on this cube (a standard number cube has #1-6) but I’m going to go with 10 numbers (don’t know how this is possible but anyway-). I’ll edit my answer if needed.

10 numbers

10 (numbers) - 2 (#8 and #10) = 8 (possible numbers)

so 8/10 or 80%

reduced fraction: 4/5

percentage: 80%

fraction: 8/10

*will edit if needed

3 0

2 years ago

A and B together complete a work in 20days. They worked together for 5days and B leaves. How long will A takes to finish the rem

andreev551 [17]

Answer:

27 days

Step-by-step explanation:

A alone, rate per day = 1/36

A and B together, rate per day = 1 / 20

Total Workdone for 5 days :

1/ 20 * 5 = 5 / 20 = 1/4

Fraction of work left :

1 - 1/4 = 3/4

3 / 4 * 36 /1

= (3 * 36) / 4

= 108 / 4

= 27 days

3 0

2 years ago

At most, how many one sixth 1 6-inch-long pieces of string can you cut from a two thirds 2 3-inch-long piece of string?

bija089 [108]

2\3 23-1\6 16
2\3-1\6=5\6
23-16=7
7 5\6

4 0

3 years ago

Question:

nikdorinn [45]

25 were on the roller coaster.
Multiplication and subtraction

7 0

3 years ago

Read 2 more answers

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

2 years ago

X +10=14 <--how does one solve this?​

X +10=14 <--how does one solve this?