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2 years ago

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A geometry teacher asked Saul to define “obtuse triangle.” Saul said that an obtuse triangle is a triangle with one interior ang

le measure greater than 90° and two interior angle measures less than 90°. Is Saul's definition valid?

1 answer:

olga_2 [115]2 years ago

7 0

It is valid.
The three angles must add up to 180. So, if one is more than 90, the other two have to be less than 90.

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Write the expanded form of h(3k-12.4)

ExtremeBDS [4]

Answer:3hk - 12.4h

Step-by-step explanation:

h(3k-12.4)

h x 3k - h x 12.4

3hk - 12.4h

3 0

2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So 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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

2 years ago

Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)

grigory [225]

Seems to be that the limit to compute is

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+2y^2}$

Consider an arbitrary line through the origin $y=mx$ , so that we rewrite the above as

$\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}$

The value of the limit then depends on the slope $m$ of the line chosen, which means the limit is path-dependent and thus does not exist.

8 0

2 years ago

Read 2 more answers

A limited-edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year

Roman55 [17]

Answer:

The required equation is $y = 18(1.15)^x$ .

Step-by-step explanation:

Consider the provided information.

The Initial value of poster = $ 18

After 1 year amount of increase = $ 20.70

With the rate of 15% = 0.15

Let future value is y and the number of years be x.

$y = 18(1.15)^x$

Now verify this by substituting x=1 in above equation.

$y = 18(1.15)^1=20.7$

Which is true.

Hence, the required equation is $y = 18(1.15)^x$ .

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2 years ago

Read 2 more answers

Ali, Ben and Clare each played a game. Clare's score was seven times Ali's score. Ben's score was half of Clare's score. Write d

Lerok [7]

Answer:

The answer is 2 : 7 : 14

8 0

2 years ago