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astraxan [27]
1 year ago
10

Help me on number 10 please.​

Mathematics
2 answers:
wel1 year ago
8 0

Answer: The answer is 32.

Step-by-step explanation:

48 <u>Divided By 3</u><em> Multiplied by 2 = 32 </em>

<em></em>

<em>So the value of N is: 32.</em>

LenaWriter [7]1 year ago
3 0

Answer:

9.) B. division

10.) 8

Explanation:

9. Which operation to do first in order of operations?

Answer: Applying BODMAS where:

[B] brackets

[O] order

[D] division

[M] multiplication

[A] addition

[S] subtraction

Maintaining order, division operation is done first. Then multiplication and then addition and so on.

10. Given that N = 48 ÷ 3 × 2

⇒ N = 48 ÷ 3 × 2

⇒ N = 48 ÷ 6

⇒ N = 8

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Tom rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a factor of 21.
Ber [7]

The probability of getting a total that is a factor of 21 is 0.22

<u>Explanation:</u>

Factor of 21 is = 1, 3, 7, 21

Out of all the factors of 21, we cannot get 21 when the two die are rolled as the maximum sum that can be obtained is 12

The possible combinations are:

(1,2) (1,6)

(2,1) (2,5)

(3,4)

(4,3)

(5,2)

(6,1)

Total outcome = 6 X 6

                      = 36

Number of outcomes that is the factor of 21 is 8

The probability of getting a total that is a factor of 21 is \frac{8}{36}

p(21) = 0.22

Therefore, the probability of getting a total that is a factor of 21 is 0.22

5 0
3 years ago
Please help. NEED THE ANSWER SOON AS POSSIBLE.
rewona [7]

Answer:

y=4x-17

Step-by-step explanation: Distribute the 4(x-5)

4x-20

Then add three to both sides

y=4x-17

6 0
3 years ago
~(25 Points)~
murzikaleks [220]
The answer is the third one
3 0
3 years ago
A sum of 6(3x+8)+32+12x in different ways
Lemur [1.5K]
First you have multiply 6 by both 3x and 8
6*3x = 18x
6*8= 48
18x + 48 + 32 + 12x
now combine like terms
18x + 12x = 30x
48 + 32 = 80
so 30x + 80 is your answer
8 0
3 years ago
If f(x)=3x^2 and g(x)=4x^3, what is the degree of (f*g)(x)?
Westkost [7]
3x^2 * 4x^3 = 12x^5

Answer is degree 5
3 0
3 years ago
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