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rosijanka [135]
2 years ago
15

I need major help now it’s an emergency

Mathematics
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

yes because together the first 2 people made 27 then you would add 8 to that

Step-by-step explanation:

You might be interested in
What equation is the equation of the line that passes through (-10,8) and (5,5)
Fofino [41]
The equation of the line is given by

y = mx + c

where m is the gradient of the line
c is where the line cuts the y-axis
x & y represent coordinates on the line.

The gradient m can be obtained as follows:

m = (5 - 8) / (5 - - 10) = (-3) / (15) = - 1/5

To obtain c, we use any known coordinate on the line and substitute it as well as the gradient in the general equation for the line.

Taking coordinates (5,5)

5 = (- 1/5)(5) + c

5 = - 1 + c

c = 6

Hence, the equation for this line is

y = -x/5 + 6
6 0
3 years ago
If a point is translated horizontally,the x-coordinate will stay the same.true or false
icang [17]
"horizonally" and "x-coordinate" are very much related.

If you start with a point and move the point horiz., the x-coordinate will change accordingly.  If the original point were (2,3) and the point is translated 3 units to the right, then the new x-coord. would be 2+3, or 5:  (5,3).
6 0
2 years ago
Read 2 more answers
Find the missing term (x^12)^5*(x^-2)^9___=(x^40)^5
ratelena [41]
(x^12)^5=x^(12*5)=x^60
(x^(-2))^9=x^(-18)
x^60*x^(-18)=x^(40) and we need to find x^a such that
x^(42)*x^a=x^(40)^5
x^a=x^(40)^5*x^(-42)
x^a=x^(200)x^(-42)
x^a=x^(200-42)
x^a=x^158
so the missing term is x^158
6 0
2 years ago
At a peace summit, seven Hatfield and nine McCoy family
katrin2010 [14]

Answer:

0.525 = 52.5% probability that the two are from different families.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, the order in which the two participants are selected is not important, so we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Desired outcomes:

1 from the Harfield family(from a set of 7).

1 from the McCoy family(from a set of 9). So

D = C_{7,1}*C_{9,1} = \frac{7!}{1!6!}*\frac{9!}{1!8!} = 7*9 = 63

Total outcomes:

2 from a set of 16. So

T = C_{16,2} = \frac{16!}{2!14!} = 120

Probability:

p = \frac{D}{T} = \frac{63}{120} = 0.525

0.525 = 52.5% probability that the two are from different families.

3 0
2 years ago
0.644×0.25 <br><br>Need the asap​
Oduvanchick [21]

Answer:

0.161

Step-by-step explanation:

The answer to 0.644 times 0.25 will end up resulting in the answer above

I'm assuming this was a simple multiplication problem? Correct me if I'm wrong on that :P

5 0
2 years ago
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