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stiks02 [169]
4 years ago
10

PLEASE HELP GUYS........................................................I'M DYING WITH ALGEBRA 1..............8TH GRADE.........

.................................SOOOOOOOOOOO HHHHHHHHAAAAAAAAAAARRRDDDDDDDDDDDD

Mathematics
1 answer:
Marat540 [252]4 years ago
8 0
24) mari=320
jen=160

25) David= 160 Aaron=55

26)Chaya= 65 Angelica=69 Jon=62

27) 8 in one group, 16 in the other
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In a homicide case 6 different witnesses picked the same man from a line up. The line-up contained 5 men. If the identifications
Vedmedyk [2.9K]

Answer:

since we have 6 witnesses and 5 men so,

therefore the probability that they would pick the same man is \frac{5}{6} =0.83333 to 5 decimal places

Step-by-step explanation:

probability is the likelihood of an event occurring or not happening.

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3 years ago
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4 years ago
4/5 of the rectangle to the left is shaded. How many extra units need to be shaded to increase this to 5/6
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6 0
3 years ago
Suppose x, y and z are integers. Prove that, if 3x−y + 5z is even, then at least one of x, y or z is even.
Elden [556K]

Given:

x, y and z are integers.

To prove:

If 3x-y+5z is even, then at least one of x, y or z is even.

Solution:

We know that,

Product of two odd integers is always odd.          ...(i)

Difference of two odd integers is always even.          ...(ii)

Sum of an even integer and an odd integer is odd.      ...(iii)

Let as assume x, y and z all are odd, then 3x-y+5z is even.

3x is always odd.          [Using (i)]

5z is always odd.          [Using (i)]

3x-y is always even.       [Using (ii)]

(3x-y)+5z is always odd.       [Using (iii)]

3x-y+5z is always odd.

So, out assumption is incorrect.

Thus, at least one of x, y or z is even.

Hence proved.

4 0
3 years ago
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