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2 years ago

Which value is equivalent to 2^6/2^3? A:2^2 B:2^3 C: 2^9 D:2^18

Mathematics

2 answers:

gavmur [86]2 years ago

5 0

I amm Answering this question so this person can get Brainly

lidiya [134]2 years ago

4 0

Answer:

2^3 or 8

Step-by-step explanation:

when you divide an exponent by an exponent, you subtract the 2 exponents

For example, 2^6/2^3=2^(6-3)=2^3

If you want to simplify 2^3, it is equal to 8

You might be interested in

If the bowl weighs 520 grams when it holds 15 tablespoons of sugar and 680 grams when it holds 26 tablespoons of sugar, what is

KATRIN_1 [288]

Answer:

1.5 grams per tablespoon

8 0

3 years ago

Roll two fair dice. Each die has six faces. A. Let A be the event that either a 3 or 4 is rolled first followed by an odd number

zhenek [66]

Answer:

A. 0.16

B. 0.58

C. No

D. Dependent

Step-by-step explanation:

The sampling space has 36 outcomes.

The event A has 6 outcomes (3,1), (3,3), (3,5), (4,1), (4,3),(4,5)

So P(A)= 6/36 = 0.16

The table in the picture shows the 15 outcomes that are not allowed, so there are 36-15 = 21 possible outcomes for B and its probability is 21/36 = 0.58

Are A and B mutually exclusive events?

No since their intersection is not empty, for example (3,1) belongs to both A and B

In order for A and B to be independent, it must happen that

P(A∩B)=P(A)P(B)

A∩B={(3,1),(3,3),(4,1)} so P(A∩B) = 3/36

whereas

P(A)P(B) = (6/36)(21/36) = (1/6)(7/12)=7/72

So A and B are not independent.

7 0

3 years ago

250 grams of sugar cost 150 dollars at Shop X. 100 grams of sugar cost 95 dollars at Shop Y. What is the difference in cost per

nata0808 [166]

<h2>Answer:</h2>

350 dollars

<h2>Step-by-step explanation:</h2>

<em>(step 1) Find the cost per kilogramme of sugar in shop X</em>

In shop X:

250 grams of sugar costs 150 dollars.

<em>Remember that;</em>

1000 grams = 1 kilogramme.

=> 250 grams will be equal to 0.25 kilogramme

Therefore (replacing 250 grams with 0.25 kilogramme) in shop X;

0.25 kilogramme of sugar costs 150 dollars

Now, if

0.25 kilogramme = 150 dollars,

=> 1 kilogramme = 150 dollars x 1 kilogramme / 0.25 kilogramme

=> 1 kilogramme = 600 dollars

Therefore, the cost per kilogramme of sugar in shop X is 600 dollars

<em>(step 2) Find the cost per kilogramme of sugar in shop Y</em>

In shop Y:

100 grams of sugar costs 95 dollars.

<em>Remember that;</em>

1000 grams = 1 kilogramme.

=> 100 grams will be equal to 0.1 kilogramme

Therefore (replacing 100 grams with 0.1 kilogramme) in shop Y;

0.1 kilogramme of sugar costs 95 dollars

Now, if

0.1 kilogramme = 95 dollars,

=> 1 kilogramme = 95 dollars x 1 kilogramme / 0.1 kilogramme

=> 1 kilogramme = 950 dollars

Therefore, the cost per kilogramme of sugar in shop Y is 950 dollars

<em>(step 3) Find the difference in cost per kilogramme of sugar between the two shops</em>

The difference is calculated by subtracting the cost per kilogramme of sugar in shop X from the cost per kilogramme of sugar in shop Y. i.e

=> cost per kilogramme of sugar in shop Y - cost per kilogramme of sugar in shop X

=> 950 dollars - 600 dollars

=> 350 dollars.

Therefore, the difference in cost per kilogramme is 350 dollars

3 0

3 years ago

who won the tournament? both teams, they ______________________ for first place. a. tie b. tied c. are tying d. had tied e. won

shutvik [7]

I feel that (a) is the right answer because both teams tie the tournament.

8 0

3 years ago

Consider M, N, and P. collinear points on MP.

Kobotan [32]

Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}$

2. Point N lies to the right from point P, then MP + PN = MN. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}$

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}$

3 0

3 years ago