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notka56 [123]
2 years ago
12

What step do u preform first when evaluating the expression 3 + (18 - 6) + 20 divided by 4

Mathematics
2 answers:
Elena L [17]2 years ago
5 0

Answer:

Step-by-step explanation:

=3+(18-6)+20÷4

=3+12+20÷4

=3+12+5

=20

use BODMAS rule (Bracket of Division, Multiplication, Addition, and Subtraction)

Lemur [1.5K]2 years ago
3 0

Answer:

20 is the answer.

Step-by-step explanation:

By using BODMAS,

→ 3 + (18 - 6) + 20 ÷ 4

→ 3 + 12 + 20 ÷ 4

→ 3 + 12 + 5 = 20

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Drag each tile to the table to multiply each row heading by each column heading
weeeeeb [17]

Answer:

multiplication by d=> 2d^3, 11d^2, -4d

multiplication by -9=>-18d^2,-99d, 36

The answers are in order from left to right.

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
What is the largest digit N for which 2345N is divisible by 6?
kotegsom [21]
Consider this option (this is not the only way):
1. according to the condition 2345N is divisible by <em>6</em>, it means, that N is even number (0;2;4;6 or 8).
2. <em>6</em>=3*2, then the number 2345N is divisible by 3 also. It means, that the number (2+3+4+5+N)=(14+N) is divisible by 3, where N={0|2|4|6|8}.
3. using items 1 and 2:
(14+4)=18; only 18 is divided by 3, it means that N=4.

answer: 4.

Note also, that only 23451; 23454 and 23457 are divided by 3. Only 2345<u>4</u> is divided by 6 (2345<u>4</u>/6=3909).
3 0
3 years ago
If a printer prints 28 copies in 1 minute how many would it print in 3 minutes and 45 seconds?
Wittaler [7]

3minutes 45 seconds = 3 45/60 = 3.75 minutes

28 copies/ 1 minute = x copies/3.75 minutes

using cross products

28 * 3.75 = 1 * x

105 = x

We can print 105 copies in 3 3/4 minutes

3 0
3 years ago
Two consecutive positive integers have a product of 6 what are the integers
umka21 [38]

Answer:

2,3

Step-by-step explanation:

2 and 3 are consecutive numbers, and when you multiply them, you get 6.

Hope this helps.

Have a nice day.

8 0
3 years ago
A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin
Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

P(\dfrac{X}{Y}) = 60\% = 0.6

P(\dfrac{X}{Y'}) = 0.01\% = 0.0001

P(Y) = 0.01

Thus, P(Y') \ will\ be = 1 - P(Y)

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

P(YX) = P(\dfrac{X}{Y}) \ P(Y)

P(YX) =0.6 \times 0.01

P(YX) =0.006

The probabilities of not involved in cheating & the evidence are present is:

P(Y'X) = P(Y')  \times P(\dfrac{X}{Y'})

P(Y'X) = 0.99  \times 0.0001 \\ \\  P(Y'X) = 0.000099

(b)

The required probability that the evidence is present is:

P(YX  or Y'X) = 0.006 + 0.000099

P(YX  or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}

P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}

P(\dfrac{Y}{X}) = 0.9838

5 0
3 years ago
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