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2 years ago

Does the table below represent a proportional relationship?

Mathematics

1 answer:

Ad libitum [116K]2 years ago

6 0

Answer:

Step-by-step explanation:

Here, we want to know if the table represents a proportional relationship

Looking at the table, if we multiply the x-terms by -4, we get the y-terms

So the answer here is that Yes, because all the ratios of y to x are equal to -4

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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)

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

$P(1\ girl) = 0.375$

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

A gold mine has two elevators, one for equipment and another for the miners. The equipment elevator descends 4 feet per second.

Ad libitum [116K]

Let $d_m,\ d_e$ represent the depth of the miner and equipments elevator, respectively. We know that equipment elevator descends with a velocity of $v_e = 4$ feet per second, and the miners one descends with a velocity of $v_m = 15$ feet per second.

If we start counting time ( $t=0$ ) when the equipment elevator begins to descend, after $t$ seconds its depth will be

$d_e(t) = v_et = 4t$

On the other hand, the miner elevator follows the same rule, but we have to use its velocity, and remember that it starts with a delay of thirty seconds:

$d_m(t) = v_m(t-30) = 15(t - 30)$

Now, we have to wait the 30 seconds of delay, and then another 14 seconds. This means that we want to know the positions of both elevators when $t = 44$ . Let's plug this value into the two equations: