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3 years ago

The graph of f(t) = 6 x 2^t shows the value of a rare coin in year t. What is the meaning of the y-intercept?

Mathematics

2 answers:

tester [92]3 years ago

4 0

Answer: The meaning y-intercept is the initial value of a rare coin is 6.

Step-by-step explanation:

Given : The graph of $f(t) = 6\times 2^t$ shows the value of a rare coin in year t.

The y- intercept is the point on Cartesian plane when the graph intersects y axis. It given the initial value of the function by substituting the value of independent variable as zero.

i.e. , we need to put t=0, we get

$f(0) = 6\times 2^0=6$

The meaning y-intercept is the initial value of a rare coin is 6.

Flauer [41]3 years ago

3 0

When the graph intersects the y-axis, it means it's x-coordinate is 0, or t = 0.
Hence, f(t) = 6 × 2^0 = 6.
y = 6 at y-intercept.

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3 years ago

How to get the answers?

beks73 [17]

By inclusion/exclusion,

$n(P' \cup Q) = n(P') + n(Q) - n(P' \cap Q)$

We have

$n(\xi) = n(P) + n(P') \implies n(P') = 28 - n(P)$

so that

$n(P' \cup Q) = (28 - n(P)) + n(Q) - 2n(P) = 28 - 3n(P) + n(Q)$

Now,

$n(\xi) = 28 \implies n(P \cup Q) = 28 - 7 = 21$

and by inclusion/exclusion,

$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$

Decompose $Q$ into the union of two disjoint sets:

$Q = (P \cap Q) \cup (P' \cap Q)$

Since they're disjoint,

$n(Q) = n(P\cap Q) + n(P'\cap Q) \implies n(Q) = n(P\cap Q) + 2n(P)$

$\implies n(P \cup Q) = n(P) + (n(P\cap Q) + 2n(P)) - n(P \cap Q)$

$\implies 21 = 3n(P)$

$\implies n(P) = 7$

From the Venn diagram, we see there are 3 elements unique to $P$ - by the way, this is the set $P \cap Q'$ - so $n(P\cap Q) = 7-3 = 4$ , and it follows that

$n(Q) = n(P\cap Q) + 2n(P) = 4 + 2\times7 = 18$

Finally, we get for (a)

$n(P' \cup Q) = 28 - 3n(P) + n(Q) = 28 - 3\times7 + 18 = \boxed{25}$

For (b), we have by inclusion/exclusion that

$n(P \cup Q') = n(P) + n(Q) - n(P \cap Q') = 7 + 18 - 3 = \boxed{22}$

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