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yKpoI14uk [10]
2 years ago
5

Review the graph.

Mathematics
2 answers:
WITCHER [35]2 years ago
6 0

Answer:

A. quadrant l

Step-by-step explanation:

Edge 2020

Helga [31]2 years ago
4 0

Answer:

a

Step-by-step explanation:

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X^5y^4+x^3y^4<br><br>x <br><br>5<br><br> y <br><br>4<br><br> +x <br><br>3<br><br> y <br><br>4
saul85 [17]

Step-by-step explanation:

x+5y×4+x×3y+4

5xy×4x+3xy+4

27xy^2

5 0
3 years ago
VERY EASY, WILL GIVE 50 POINTS FOR CORRECT ANSWER ASAP AND WILL GIVE BRAINLIEST.
Arada [10]
The answer is d. You are welcome
8 0
3 years ago
Abiodun and taiwo share 735 naira between them so that abiodun get 95naira more than taiwo find how much money each gets
DerKrebs [107]

Answer:

Abiodun gets - <u>415 naira</u>

Taiwo gets -<u> 320 naira</u>

Step-by-step explanation:

Let us consider  x and y as Abiodun and Taiwo,

Therefore , acccording to the question , equation formed will be -

 x+y=735 naira ------ 1

and

x=y+95

x-y=95 ------2

From equation 1 and 2, we get

   2x=830

x=\frac{830}{2}

x=415 naira (that is Abiodun share)

now , Calculating Taiwo share from equation 1

x+y=735\\415+y=735

y=735-415

y = 320 naira (that is Taiwo share )

Therefore , Share of Abiodun = 415 naira

   Share of Taiwo = 320 naira

5 0
3 years ago
Lena won a charity raffle. Her prize will be randomly selected from the 9 prizes shown below. The prizes include 7 rings, 1 came
pickupchik [31]

Answer:

(b) Find the odds in favor of Lena winning a headset ​

Step-by-step explanation:

4 0
3 years ago
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
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