Ask question

Ask question

All categories

xxTIMURxx [149]

2 years ago

11

A bag contains 15 green, 18 yellow, and 16 orange balls. One ball is randomly selected. To the nearest percent, what is the prob

ability of the event? Drag and drop the correct value into the box.
33%
31%
37%
42%

1 answer:

scZoUnD [109]2 years ago

4 0

37%
15+18+16=49
49/.37=18.13
Your biggest number is 18 so that means 37% would be the answer.
Hope it helps

You might be interested in

HELP ASAP!! DUE TOMORROW!!!!!! NUMBER #6!!

Rashid [163]

Its B. I cam tell by looking.

6 0

3 years ago

alukav5142 [94]

Answer:

C

Step-by-step explanation:

This question can simply be answered by the process of elimination. All of the other answers have numbers behind the first number of the decimal. when you are rounding to the nearest tenth, you will have no numbers behind the decimal except 1. They will look like this:

11.70

23.30

45.60

So, since we are looking for the nearest tenth, we would choose C because it is the only one that is rounded.

8 0

2 years ago

10^-3 mm in inches??

Anarel [89]

3.9

it has to be 20 characters long so hi

6 0

2 years ago

Read 2 more answers

Korie randomly selects a number from the numbers shown below. 54, 60, 11, 0, 5, 7, 20 What is the probability that the number sh

grandymaker [24]

Answer:

there is a greater chance that she selects a even number because there are 4 even numbers and 3 odd numbers

8 0

2 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

2 years ago