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2 years ago

PLZ HELP 500 POINTS. PLZ SHOW YOUR WORK + BRAINLIEST

Mathematics

2 answers:

sesenic [268]2 years ago

7 0

Answer:

A) She should buy the

Step-by-step explanation:

Part A:

For question A, we find the areas of both plots and see which one is greater.

For the first one, we do 3/8 * 2/3

$\frac{3}{8} * \frac{2}{3} = \frac{6}{24} = \frac{1}{4}$

The area of the first plot is 1/4 square miles

For the second one, we do 2/5 * 2/5

$\frac{2}{5} *\frac{2}{5} = \frac{4}{25}$

The area of the second plot is 4/25 square miles

We have 1/4 and 4/25.

To see which one is bigger, we make them have the same denominator, also known as the LCM (Least Common Multiple).

LCM(4,25) = 100

Now we make 1/4 have a denominator of 100

4 times what is 100?

25, so the numerator is 25 and the denominator is 100.

We get: 25/100

Now we make 4/25 have a denominator of 100

25 times what is 100?

4, so the numerator is 4 * 4 = 16 and the denominator is 100.

We get: 16/100

Now we can see which is bigger:

25/100 or 16/100

25/100 is bigger

Therefore, Miguel's mother should buy the plot with dimensions 3/8 and 2/3.

Part B

For part B, I'm not sure which plot of land she is going to pay for, so I will answer both.

We answered this in Part A already.

The 3/8 and 2/3 plot was 1/4 square miles and the 2/5 by 2/5 plot was 4/25 square miles.

Plot 1:

1/4 square miles

It said that it costs $750,000 for a square mile

This plot is 1/4 square miles, so we do

$$750,000 * \frac{1}{4} = \frac{750,000}{4} = 187,500$

The cost for plot 1 is $187,500

Plot 2:

4/25 square miles

It said that it costs $750,000 for a square mile

This plot is 4/25 square miles, so we do

$750,000*\frac{4}{25} = \frac{3,000,000}{25} = 120,000$

The cost for plot 1 is $120,000

So for the 3/8 by 2/3 plot, it cost $187,500 and for the 2/5 by 2/5 plot costs $120,000

ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘᴇᴅ!

ʜᴀᴠᴇ ᴀ ɴɪᴄᴇ ᴅᴀʏ!

Jet001 [13]2 years ago

4 0

Answer: don't know sorry

Step-by-step explanation:

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3 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