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2 months ago

A circle has diameter of 11cm

Mathematics

1 answer:

Zina [12.3K]2 months ago

5 0

Since the diagonal's length ( $Diagonal= 9.9cm$ ) is shorter than the circle's diameter (11 cm), the square can be accommodated within the circle without making contact with its boundary.

Detailed explanation:

Consider that a circle has a diameter of 11 cm and a square with a side length of 7 cm. We need to apply Pythagoras’ Theorem to demonstrate that the square will fit within the circle without touching its boundary. Let's analyze:

One knows that to allow a square to fit within a circle without touching its edge, the diagonal measurement of the square must remain smaller than the circle's diameter. To derive the diagonal's length, use Pythagoras' theorem:

$Hypotenuse ^2 = Perpendicular^2+Base^2$

For a square, $Perpendicular = base = side$

⇒ $Diagonal^2 = 2(side)^2$

⇒ $Diagonal= \sqrt{2}(side)$

⇒ $Diagonal= \sqrt{2}(7)$

⇒ $Diagonal= 9.9cm$

Since the diagonal's length ( $Diagonal= 9.9cm$ ) is shorter than the circle's diameter (11 cm), it can be confirmed that the square fits inside the circle without touching the edge of the circle.

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According to a survey by Bankrate, of adults in the United States save nothing for retirement (CNBC website). Suppose that adult

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Complete Question

The complete question appears in the first uploaded image

Answer:

a) Yes, selecting 15 corresponds to a binomial experiment

d) $P(r = 15) = 3.2768 *10^{-11}$

Step-by-step explanation:

Regarding question a:

For an experiment to qualify as binomial

the trials have to be independent

each trial must yield one of two possible outcomes

Given that the selection of 15 individuals is random, we ascertain that the trials are independent and the outcomes are “either the individual saves for retirement or does not save for retirement.”

Therefore, we conclude that the selection of 15 people at random is indeed a binomial experiment.

In question b:

The probability that all selected adults do not save for retirement is mathematically modeled as

$P(r = n) = ^nC_r * p^r * q^{n-r}$

Here C signifies combination

r = 15 implies all selected adults

n refers to the population size equating to 15

From the problem, p = 0.20

and q can be calculated as

=> $q = 1 - p$

Thus

$P(r = 15) = ^{15}C_{15} * p^{15} * q^{15-15}$

$P(r = 15) = 3.2768 *10^{-11}$ Regarding question c:

The probability that exactly five of the selected adults do not save for retirement is mathematically modeled as

$P(r = 5) = ^{15} C_5 * (0.20)^5 * (0.80)^{15}$

$P(r = 5) = 0.1032$

In relation to question d:

The probability that at least one of the selected adults opts not to save for retirement can be mathematically expressed as

$P(r \ge 1 ) = 1 - P (r = 0 )$

$P(r \ge 1 ) = 1 - [ ^{15} C _ 0 * (0.20)^{0} * (0.80 )^{15}]$

$P(r \ge 1 ) = 1 - 0.0352$

$P(r \ge 1 ) = 0.9648$

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A function and its inverse, when graphed, exhibit symmetry about the line y = x. Without viewing your graphs, I can't specify if g or h is the inverse. Identify the function that reflects f over the diagonal line y = x, which passes through the origin.

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3 months ago

On a coordinate plane, 2 exponential fuctions are shown. Function f (x) decreases from quadrant 2 into quadrant 1 and approaches

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Answer:

$g(x)=6(3)^x$

Step-by-step explanation:

We are provided with the information that

$f(x)=6(\frac{1}{3})^x$

Function f is decreasing as it moves from quadrant 2 to quadrant 1, tending towards y=0

It intersects the y-axis at the point (0,6) and goes through (1,2).

Function g(x) also approaches y=0 in quadrant 2, but increases in quadrant 1.

It passes through (-1,2) and crosses the y-axis at (0,6).

Reflection across y-axis:

The transformation rule is identified as

$(x,y)\rightarrow (-x,y)$

Applying this rule, we derive

$g(x)=6(\frac{1}{3})^{-x}=6(3)^x$

Then, by substituting

$x^{-a}=\frac{1}{x^a}$

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Consequently, $g(x)=6(3)^x$ holds true.

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