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Lelu [443]
3 years ago
6

Help please ((: ughhh

Mathematics
1 answer:
Dvinal [7]3 years ago
4 0

1/2............................................ I hope this helps



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marin [14]
2.57=3
+0.14=0
———
2.71=3

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Use prime factorization to find gcf of 57 and 27
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Gcf (57,27) = <span><span><span>57×27/</span><span>lcm(57,27)</span></span>=<span>1539/513</span></span> = 3
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I'M TERRIBLE AT MATH SOMEONE HELP, WILL GIVE BRAINLIEST
jonny [76]

Answer:

D) -1.8, -1 1/2 , 0.3 , 3/4 , 1

Step-by-step explanation:

Change all of the given terms to either fractions or decimals. Generally decimals are easier:

3/4 = 0.75

-1.8 = -1.8

1 = 1.0

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Order from least to greatest:

-1.8 , -1.5, 0.3 ,  0.75 , 1.0

or

-1.8, -1 1/2 , 0.3 , 3/4 , 1

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5 0
3 years ago
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Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

7 0
3 years ago
What set of refections would carry trapezoid ABCD onto itself?
BaLLatris [955]
Hello!

Trapezoid ABCD is shown. A is at negative 5, 1. B is at negative 4, 3. C is at negative 2, 3. D is at negative 1, 1.
A) x-axis, y=x, y-axis, x-axis
b) x-axis, y-axis, x-axis
c) y=x, x-axis, x-axis 
d) y-axis, x-axis, y-axis, x-axis
The best answer is C180 rotation wud take that point to 4th quadrant
reflection in x-axis takes that to 1st quadrant
<span>reflection in y-axis brings it back to 2nd quadrant again. So, the sequence of transformations will  bring A back to where it started
</span>
Hope this Helps! :)
5 0
3 years ago
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