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3 years ago

4x-7y=-8 and 3x+2y=-6 solve for x and y

Mathematics

1 answer:

Mariulka [41]3 years ago

6 0

Step-by-step explanation:

4x-7y= -8

or, 4x= 7y-8

or, x = 7y-8 ...........(i)

3x+ 2y =-6

or, 3x= -(2y+6)

or, x= -(2y+6) ........... (ii)

from (i) and (ii),

7y-8 = -(2y+6)

4 3

or, 21y-24 = -8y-24

or,

y =

then , Let's put the value of y in (i) or (ii)

x =

sorry I'm tired

so, i can't complete this

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g natasha is in a class of 30 students that selects 4 leaders. How many ways are there to select the 4 leaders so that natasha i

Papessa [141]

Answer:

<h2>3,654 different ways.</h2>

Step-by-step explanation:

If there are 30 students in a class with natasha in the class and natasha is to select four leaders in the class of which she is already part of the selection, this means there are 3 more leaders needed to be selected among the remaining 29 students (natasha being an exception).

Using the combination formula since we are selecting and combination has to do with selection, If r object are to selected from n pool of objects, this can be done in nCr number of ways.

nCr = n!/(n-r)!r!

Sinca natasha is to select 3 more leaders from the remaining 29students, this can be done in 29C3 number of ways.

29C3 = 29!/(29-3)!3!

29C3 = 29!/(26!)!3!

29C3 = 29*28*27*26!/26!3*2

29C3 = 29*28*27/6

29C3 = 3,654 different ways.

This means that there are 3,654 different ways to select the 4 leaders so that natasha is one of the leaders

5 0

3 years ago

The histograms below show the mass (in kilograms) of the offensive linemen on the University of Mount Union and the University o

Sedbober [7]

Answer:

Option C is correct.

None of the options is true.

Step-by-step explanation:

The histograms for the question is missing, so, it was obtained online and is attached to this solution of the question.

To check which statements are true, we first find the average using the midpoint technique.

The midpoint for each of the ranges is presented below

110 to 120 = 115

120 to 130 = 125

130 to 140 = 135

140 to 150 = 145

150 to 160 = 155

Sum of weights for Alabama linemen using the midpoint technique

= (125×4) + (135×8) + (145×4) + (155×2)

= 2470 kg

Average = (2470) ÷ (4+8+4+2)

= 137.22 kg

Sum of weights for Mount Union linemen

= (115×5) + (125×4) + (135×6) + (145×2)

= 2,175 kg

Average = (2,175) ÷ (5+4+6+2)

= 127.94 kg

So, it is evident that the average weight of Mount Union linemen isn't more than that of Alabama linemen.

Statement 2

All of the Alabama linemen have more mass than all of the Mount Union linemen

This is evidently not true because from the histogram, some linemen from Alabama have masses between 120 and 130, 130 and 140 & there are linemen from Mount Union that have masses between 130 and 140 & 140 and 150, so, it isn't right to say that all of the Alabama linemen have more mass than all of the Mount Union linemen.

So, none of the options is correct, option C.

Hope this Helps!!!