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3 years ago

Guy's y'all plzz answer this who knows about it...and also exam is near

Mathematics

1 answer:

Yanka [14]3 years ago

7 0

Answer:

Step-by-step explanation:

Okay so I will explain the first one.

2 to the power of 4 = 2x2x2x2

2 to the power of 6 = 2x2x2x2x2x2

2 to the power of 4 = 2x2x2x2

2^4 = 16

2^6 = 64

When you are multiplying powers you add.

When you subtract you takeaway.

So you would do

4+6+4

= 14

Meaning 2 to the power of 14

2 $x^{14}$

So what is 2 to the power of 14.

You would do

2x2x2x2x2x2x2x2x2x2x2x2x2x2

The answer is 16384

You can also use a calculator

GOOD LUCK FOR YOUR TEST!

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Can you please help me....Thank you!

PIT_PIT [208]

Answer:

mean=7

median=8

range=9

mode=3

mean=15

mode=17

range=14

median=14

mean=11

mode=none

range=10

median=4.5

mean=6

mode=4

range=13

median=4

Step-by-step explanation:

there you go, I think that's right but it's been awhile since I've had to do mode mean median and range

4 0

2 years ago

The total area of the figure below is (blank) units look at picture to solve

Tpy6a [65]

9x4=36 squared units

3 0

3 years ago

I need help to this question

dlinn [17]

The range would be $14.

To find the range, you would take the highest value and subtract it by the lowest value:
29 - 43 =
14

Hope this helped! :)

5 0

2 years ago

Read 2 more answers

A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taken out of the box, one at a

USPshnik [31]

Answer:

$Probability = 0.504$

Step-by-step explanation:

Given

$Gems= 35$

$Real = 10$

$Fake = 25$

Required

Determine the probability of selecting two fakes

The probability can be represented as thus: $P(Fake\ and\ Fake)$

Using the following probability formula, we have:

$P(Fake\ and\ Fake) = P(Fake) * P(Fake)$

Each probability is calculated by dividing number of fakes by total number of gems:

$P(Fake\ and\ Fake) = \frac{25}{35} * \frac{25-1}{35-1}$

The minus 1 (-1) represent the numbers of fake and total gems left after the first selection

$P(Fake\ and\ Fake) = \frac{25}{35} * \frac{24}{34}$

$P(Fake\ and\ Fake) = \frac{5}{7} * \frac{12}{17}$

$P(Fake\ and\ Fake) = \frac{60}{119}$

$P(Fake\ and\ Fake) = 0.504$

<em>Hence, the required probability is approximately 0.504</em>

7 0

3 years ago

You set up a hot chocolate stand out of football game the cost of your supplies and $75 you charge $.50 for each cup of hot choc

Elodia [21]

P = .50c - 75

You need work your cost into your total profit, $75, which was spent. Since that money was lost, we can say it is -$75. You make 50¢ per cup of hot choclate, which is $0.50. Putting that all together, you can make the function P = .50c - 75.

4 0

3 years ago

Guy's y'all plzz answer this who knows about it...and also exam is near​

Guy's y'all plzz answer this who knows about it...and also exam is near