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3 years ago

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The statue or liberty is 305 feet tall measuring from the ground to the tip of the torch if the model is 1/100 the actual size o

f the staute of liberty, how tall is the model?

1 answer:

sergejj [24]3 years ago

8 0

Given:

Height of statue or liberty from the ground = 305 feet

Model is $\dfrac{1}{100}$ the actual size of the statue of liberty.

To find:

The height of the model of statue.

Solution:

According to the question,

Model = $\dfrac{1}{100}$ of the actual size

$=\dfrac{1}{100}\times 305$

$=3.05$

Therefore, the height of the model of statue is 3.05 feet.

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Consider the graph of the sixth-degree polynomial function f.

vekshin1

Answer:

b = 1, c = -1 and d = 4

Step-by-step explanation:

To solve this question the rule of multiplicity of a polynomial is to be followed.

If the multiplicity of a polynomial is even at a point, graph of the polynomial will touch the x-axis.

If the multiplicity of the polynomial is odd, graph will cross the x-axis at that point.

From the graph of function 'f',

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Since, graph of the function 'f' crosses x-axis at x = 1 and x = 4, multiplicity will be odd and touches the x-axis at x = -1 multiplicity will be even.

So the function will be,

f(x) = (x - 1)[x - (-1)]²(x - 4)³

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The diagonals of a rhombus are 14 and 48cm. Find the length of a side of the rhombus.

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3 years ago

3/5 divided by 4

Vesna [10]

$\bold{Hey\ there!}$

$\bold{\frac{3}{5}\div\frac{1}{4}}$

$\bold{1\times3=3\ ;\ 5\times4=20}$
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$\bold{20\rightarrow denominator}$
$\boxed{\boxed{\bold{Answer:\frac{3}{20}}}}\checkmark$

$\bold{Good\ luck\ on\ your\ assignment\ and\ enjoy\ your\ day!}$

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3 years ago

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites. Assume you obtain a r

kicyunya [14]

Answer:

a) 0.2316 = 23.16% probability that 0 carry intestinal parasites.

b) 0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

Step-by-step explanation:

For each trout, there are only two possible outcomes. Either they carry intestinal parasites, or they do not. Trouts are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You know that in a specific population of rainbow trout 15% of the individuals carry intestinal parasites.

This means that $p = 0.15$

Assume you obtain a random sample of 9 individuals from this population:

This means that $n = 9$

a. Calculate the probability that __ (last digit of your ID number) carry intestinal parasites.

Last digit is 0, so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

0.2316 = 23.16% probability that 0 carry intestinal parasites.

b. Calculate the probability that at least two individuals carry intestinal parasites.

This is

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.15)^{0}.(0.85)^{9} = 0.2316$

$P(X = 1) = C_{9,1}.(0.15)^{1}.(0.85)^{8} = 0.3679$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.2316 + 0.3679 = 0.5995$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5995 = 0.4005$

0.4005 = 40.05% probability that at least two individuals carry intestinal parasites.

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3 years ago