If x2 = 30, what is the value of x?

Elliott is counting the change from a class fundraiser. He has half as many quarters as nickels, and he has 7 fewer dimes than n

Akimi4 [234]

Answer: He should have translated the scenario as value of all dimes = value of all

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A diver dives to 50 feet in 10 minutes, stays there for 30 minutes, and then resurfaces in 10 minutes. Click on the graph until

omeli [17]

Answer:

50x5

Step-by-step explanation:

8 0

2 years ago

Problem 2.2.4 Your Starburst candy has 12 pieces, three pieces of each of four flavors: berry, lemon, orange, and cherry, arrang

kkurt [141]

Answer:

a) P=0

b) P=0.164

c) P=0.145

Step-by-step explanation:

We have 12 pieces, with 3 of each of the 4 flavors.

You draw the first 4 pieces.

a) The probability of getting all of the same flavor is 0, because there are only 3 pieces of each flavor. Once you get the 3 of the same flavor, there are only the other flavors remaining.

b) The probability of all 4 being from different flavor can be calculated as the multiplication of 4 probabilities.

The first probability is for the first draw, and has a value of 1, as any flavor will be ok.

The second probability corresponds to drawing the second candy and getting a different flavor. There are 2 pieces of the flavor from draw 1, and 9 from the other flavors, so this probability is 9/(9+2)=9/11≈0.82.

The third probability is getting in the third draw a different flavor from the previos two draws. We have left 10 candys and 4 are from the flavor we already picked. Then the third probabilty is 6/10=0.6.

The fourth probability is getting the last flavor. There are 9 candies left and only 3 are of the flavor that hasn't been picked yet. Then, the probability is 3/9=0.33.

Then, the probabilty of picking the 4 from different flavors is:

$P=1\cdot\dfrac{9}{11}\cdot\dfrac{6}{10}\cdot\dfrac{3}{9}=\dfrac{162}{990}\approx0.164$

c) We can repeat the method for the previous probabilty.

The first draw has a probability of 1 because any flavor is ok.

In the second draw, we may get the same flavor, with probability 2/11, or we can get a second flavor with probability 9/11. These two branches are ok.

For the third draw, if we have gotten 2 of the same flavor (P=2/11), we have to get a different flavor (we can not have 3 of the same flavor). This happen with probability 9/10.

If we have gotten two diffente flavors, there are left 4 candies of the picked flavors in the remaining 10 candies, so we have a probabilty of 4/10.

For the fourth draw, independently of the three draws, there are only 2 candies left that satisfy the condition, so we have a probability of 2/9.

For the first path, where we pick 2 candies of the same flavor first and 2 candies of the same flavor last, we have two versions, one for each flavor, so we multiply this probability by a factor of 2.

We have then the probabilty as:

$P=2\cdot\left(1\cdot\dfrac{2}{11}\right)\cdot\left(\dfrac{9}{10}\cdot\dfrac{2}{9}\right)+\left(1\cdot\dfrac{9}{11}\cdot\dfrac{4}{10}\cdot\dfrac{2}{9}\right)\\\\\\P=2\cdot\dfrac{36}{990}+\dfrac{72}{990}=\dfrac{144}{990}\approx0.145$

5 0

2 years ago

Jessica is measuring two line segments. The first line segment is 30 cm long. The secon line segment is 500mm long. In centimete

igor_vitrenko [27]

Well 500 mm = 50 cm so if you add both segments together you will get a 50+30=80cm segment

7 0

2 years ago

Jonah was standing at an elevation somewhere between Negative 1 and one-half and Negative 2 and one-third meters with regards to

Agata [3.3K]

Options :

A number line going from negative 3 to positive 3 in increments of 1.

1 and StartFraction 5 Over 6 EndFraction meters

Negative 2 and StartFraction 3 Over 6 EndFraction meters

2 and StartFraction 3 Over 6 EndFraction meters

Negative 1 and StartFraction 5 Over 6 EndFraction meters

Answer:

Negative 1 and StartFraction 5 Over 6 EndFraction meters

Step-by-step explanation:

Jonah's position range :

Between - 1 1/2 meters and - 2 1/3 meters with respect to sea level

With regards to sea level and the range of position given ; Jonah's position will be below sea level, that is Jonah's position cannot be a positive location on the number line, thus all options with positive values are incorrect.

Hence Jonah's position will be any number on a number line located in between :

-2 1/3 and - 1 1/2

That is

-2 1/3 ≤ Jonah's location ≤ - 1 1/2

-2 1/2 ( this is less than - 2 1/3) (incorrect)

-1 5/6 lies in between the given range and is thus a possible position.

7 0

2 years ago