All categories

3 years ago

URGENT!!!! PLEASE HELP NOW!!! WHO EVER GIVES THE CORRRECT ANSWER WILL GET BRAINLIEST!!!

Mathematics

2 answers:

sladkih [1.3K]3 years ago

3 0

Answer:

The correct ans is..... ( which i believe )

3rd option

Hope this helps...

Pls mark my ans as brainliest

If u mark my ans as brainliest u will get 3 extra points

harkovskaia [24]3 years ago

3 0

Answer:

3rd option

Step-by-step explanation:

because 2/5 of 40 is equal to 16, and thats the equivelent of the pants sold.

You might be interested in

Suppose that P is the point on the unit circle obtained by rotating the initial ray counterclockwise through 330 degrees. Find t

Flauer [41]

Answer:

Reference angle will be 30°

sin(330°) will be equal to $\frac{-1}{2}$

cos(330°) will be equal to 0.866

Step-by-step explanation:

We have given angle is 330°

As the angle $330^{\circ}$ is in forth quadrant

We know that in forth quadrant the reference angle is given by

Reference angle $=360-angle$

As the angle is 330°

So the reference angle will be $=360-330=30^{\circ}$

Now we have to find the value of

sin(330°) $=sin(2\pi -30)=-sin(30)=\frac{-1}{2}$

And now $cos(330^{\circ})=cos(2\pi -30)=cos30=0.866$

4 0

3 years ago

Emily is entering a bicycle race for charity her mother pledges $0.40 every 0.25 miles she bikes only bikes 15 miles how much wi

ladessa [460]

The answer is 24 dollars

6 0

3 years ago

Can you help me with number 6? <br> Confused abit <br> Please

Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.

To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.

First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is

$\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}$

Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is

$\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}$

Third question:

To get exactly one six, it can either be the first, second or third roll.

In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.

In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus

$\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}$

And since there are three of these combinations, The answer is

$3\frac{5^2}{6^3}$

Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is

$1 - \frac{5^3}{6^3}$