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olga55 [171]
3 years ago
14

Suppose B is a proper subset of C,

Mathematics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

Maximum: 7

Minimum: 0

Step-by-step explanation:

A proper subset B of a set C, denoted B\subset C, is a subset that is strictly contained in C and so necessarily excludes at least one member of C.

This means that the number of elements in B must be at least 1 less than the number of elements in C. If the number of elements in C is 8, then the maximum number of elements in B can be 7.

The empty set is a proper subset of any nonempty set. Hence, the minimum number of elements in B can be 0.

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A couple plans to have 3 children, what is the probability that atleast two will be boys?
Olin [163]

1/4 or 25 percent because there is a 1/2 chance to get a boy or girl so the probability that will be boys us 1/2 x 1/2= 1/4

6 0
3 years ago
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Answer the photo below thanks <br> Explain how you got it
inysia [295]
The answer is 488.
22*17=374
12*19=228
228/2=114
374+114=488
7 0
3 years ago
Me ayudan hacer este ejercicio con los pasos
QveST [7]

Answer:

A) 21/7, or 3

B) 36/5, or 7 1/5

C) 35/25, or 1 15/25, or 1 3/5

D) 54/12, or 4 6/12, or 4 1/2

Step-by-step explanation:

8/7 plus 13/7. You add the numerators (the numbers on top) together. That makes 21. The denominator (the numbers on bottom) stays the same. So it would be 21/7. 7 fits into 21, 3 times.

8/7 más 13/7. Agrega los numeradores (los números en la parte superior) juntos. Eso hace 21. El denominador (los números en la parte inferior) permanece igual. Entonces sería 21/7. 7 encaja en 21, 3 veces.

8/7 plus 13/7. Vous ajoutez les numérateurs (les chiffres du haut) ensemble. Cela fait 21. Le dénominateur (les chiffres du bas) reste le même. Donc, ce serait 21/7. 7 correspond à 21, 3 fois.

8/7 plus 13/7. Quarum numeratores addere (supra de numero) una. 21. Quod facit denominator est (per numeros in fundo) manebit. Ita esset 21/7. Vicium, in VII XXI, III tempora.

6 0
3 years ago
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A pharmacist is filling small empty bottles with cough syrup from a larger bottle. He can solve the inequality to find the numbe
andrey2020 [161]

Let us assume capacity of large bottle = 1000 ml.

And capacity of each small bottle = 25 ml.

Reaming milliliters in large bottle = 250 milliliters.

Number of bottles = x bottles.

Total capacity of large bottle - capacity of a small bottle × number of bottles > Reamining milliliters in large bottle.

Therefore, we can setup an inequality.

1000 - 25 × x > 250.

1000 -25x >250.


7 0
2 years ago
Find two whole numbers with a sum of 15 and a product of 54
likoan [24]
A+b=15----(1)
a*b=54----(2)

(1) a=15-b ----(3)

put(3) in(2)

(15-b)*b=54
15b-b^2=54
b^2-15b+54=0
(b-9)(b-6)=0
so b=9 or b=6

if b=9 then a=15-9=6
if b=6 then a=15-6=9

answer two numbers are 6 and 9
3 0
3 years ago
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