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madam [21]
3 years ago
8

Daniel thinks of a number and subtracts 2

Mathematics
2 answers:
Tatiana [17]3 years ago
7 0

Answer:

the number is 126

Step-by-step explanation:

1) multiply each side by 32: (n-2) / 32 = 4

2) subtract 2 from each side: n-2 = 128

3) n = 126

Keith_Richards [23]3 years ago
7 0

Answer:

130

Step-by-step explanation:

So basically this number, minus 2, divided by 32 is 4, so lets write the unknown number as x.

So, (x-2)/32 = 4.

First multiply both sides by 32 to get rid of the denominator.

(32)(x-2)/32 = 4(32)          x-2 = 128

Now the -2 remains to figure out x, so you add 2 to both sides to get rid of the -2.

So, x-2+2 = 128+2,    -2 +2 is 0, and 128 +2 is 130, so you have x = 130.

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PLEASE ANSWER ASAP !!!!! WILL GIVE BRAINLIEST
Licemer1 [7]

Answer:

4 ≤ t ≤ 8

Step-by-step explanation:

Look at the graph and see if between the numbers whether the graph is going up or down.

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3 years ago
A student ran out of time on a multiple choice exam and randomly guessed the answers for 2 problems. Each problem had 5 answer c
frozen [14]
So if there are 5 choices and only one is correct, there is a 1/5 chance of being correct. since there was 2 questions, you need to multiply 1/5 by 1/5 to get 2/25 of getting both questions correct.

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3 years ago
The school you want to go to has an acceptance rate of 15%. That means that 15% of the students who apply to the school get in.
lisov135 [29]
12,000 Students Applied...

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4 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
erma4kov [3.2K]

Answer:

\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

Step-by-step explanation:

Given

5 tuples implies that:

n = 5

(h,i,j,k,m) implies that:

r = 5

Required

How many 5-tuples of integers (h, i, j, k,m) are there such thatn\ge h\ge i\ge j\ge k\ge m\ge 1

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed:  This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

^{n}C_r => ^{n + 4}C_5

^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}

^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}

Expand the numerator

^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}

^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}

<u><em>Solved</em></u>

6 0
3 years ago
Solve for angel GLN<br><br> please help asap, need help
likoan [24]

Answer:

119

Step-by-step explanation:

\mathrm{Supplementary\:angles\:add\:up\:to}\:180^{\circ }

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\mathrm{So\:}\angle \:GKO+\angle \:KNL=180

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\angle \:GKO=119

6 0
3 years ago
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