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steposvetlana [31]
2 years ago
15

Assuming that boy and girl babies are equally​ likely, what would be​ Kathy's probability of having if she were to have four chi

ldren​ altogether?
Mathematics
1 answer:
Mars2501 [29]2 years ago
3 0
It would be 1/2 of a chance
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Would like to see the step-by-step how to solve this!
tiny-mole [99]

The solutions to the respective equations shown in the image are 220, 41, -52, 1.385 and 11

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

a) 23 + (-16) + 42 * 5 - (-3)

= 23 - 16 + 210 + 3 = 220

b) -6(12 - 15) + 23

= -6(-3) + 23 = 18 + 23

= 41

c) -50 + (-10) + (5 - 3)4

= -50 - 10 + 2(4) = -50 - 10 + 8

= -52

d) -4.5 (-0.53) + (-1)

= 2.385 - 1

= 1.385

e) 5 - 2 + 8

= 3 + 8

= 11

The solutions to the respective equations shown in the image are 220, 41, -52, 1.385 and 11

Find out more on equation at: brainly.com/question/2972832

#SPJ1

8 0
1 year ago
At the city museum, child admission is $5.70 and adult admission is $9.80 . On Sunday, four times as many adult tickets as child
Ket [755]
5.70c + 9.80a = 1526.60
a = 4c

5.70c + 9.80(4c) = 1526.60
5.70c + 39.20c = 1526.60
44.90c = 1526.60
c = 1526.60/44.90
c = 34 <=== child tickets sold

a = 4c
a = 4(34)
a = 136 ...adult tickets sold
4 0
2 years ago
I have attached a picture of the question.​
sleet_krkn [62]
Its same i think

100-40=60
130-70=60
8 0
2 years ago
-3/10x (-0.1) can you please help ASAP
Gre4nikov [31]

\frac{ - 3}{10}  \times ( - 0.1)

\frac{ - 3}{10}  \times  \frac{ - 1}{10}

<u> </u><u> </u><u> </u><u>–</u><u>3</u><u> </u><u> </u><u> </u><u>×</u><u> </u><u> </u><u> </u><u>–</u><u>4</u><u> </u><u> </u><u> </u><u> </u><u> </u>

10

\frac{12}{10}

\frac{6}{5} answer

<h3>HOPE THIS HELPS :)</h3>
8 0
2 years ago
The sophomore class is planning to sell school mascot puppets to help cover the cost of a class trip. A local supplier will char
kifflom [539]

Answer:

755 or more

Step-by-step explanation:

The profit is the difference between revenue and costs. We want the profit to be $2000 or more, and we have both fixed and variable costs.

Let x represent the number of puppets sold. Then the costs are ...

... 76.25 + 2.25x

The revenue is 5x.

The above-described relationship can then be written as

... 5x -(76.25 +2.25x) ≥ 2000

... 2.75x ≥ 2076.25 . . . . . add 76.25, collect terms

... x ≥ 2076.25/2.75 . . . . divide by the coefficient of x

... x ≥ 755

755 or more puppets must be sold to earn $2000 or more.

8 0
3 years ago
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