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4 years ago

A debt greater than 20$ and less than 30$

Mathematics

1 answer:

Vsevolod [243]4 years ago

5 0

Answer:

A debt greater than 20$ and less than 30$ is 25$

Step-by-step explanation:

A brainliest would be nice please :D

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Jamie invests $1,200 in an account that earns 3% interest. Assuming continuous compounding, how much is in his account after 7 y

goblinko [34]

Answer:

The amount in account after 7 years of investment is $1478.4

Step-by-step explanation:

Given as :

The principal invested in account by Jamie = p = $1200

The rate of interest = r = 3% compounded annually

The Time period of investment = t= 7 years

Let The Amount in Jamie account = $ A

For continuous compounding

Amount = Principal × $e^{r \times time}$

Or, A = p × $e^{r \times time}$

Or, A = $1200 × $e^{0.03 \times 7}$

Or, A = $1200 × $2.71^{0.03 \times 7}$

Or, A = $1200 × 1.232

∴ A = $1478.4

So, The amount after continuous compounding = A = $1478.4

Hence , The amount in account after 7 years of investment is $1478.4 Answer

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4 years ago

What digit is in the ones place of the 357th even number?

wlad13 [49]

(357) the 3 is in the hundredth place, the 5 is in the tens place and the 7 is in the ones place

8 0

4 years ago

Tina cut 3 wood boards each into 4 equal pieces she used 3 of the pieces for a craft. What fraction of the boards does she have

kumpel [21]

Answer:

$2\frac{1}{4}$ .

Step-by-step explanation:

Given:

Tina cut 3 wood boards each into 4 equal pieces.

She used 3 of the pieces for a craft.

Question asked:

What fraction of the boards does she have left ?

Write this fraction as a mixed number.

Solution:

As she cuts 3 wood board each into 4 equal pieces :-

Total number of equal pieces = $3\times4=12\ pieces$

Pieces of wood used = 3

Remaining pieces = 12 - 3 = 9

<h3>Now, we will write these 9 pieces in terms of number of wood boards.</h3>

By unitary method:

4 pieces constitute = 1 wood board

1 piece constitute = $\frac{1}{4} \ wood\ board$

9 pieces constitute = $\frac{1}{4} \times9=\frac{9}{4} \ wood\ board$

Now, convert it into mixed fraction:

$\frac{9}{4} =2\frac{1}{4} \ wood\ board$

Therefore, fraction of the boards she have left is $2\frac{1}{4}$ .

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3 years ago

The square root of 125 is between what two numbers

Otrada [13]

The square root of 125 is in between 11 and 12. It is 11.1803.

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4 years ago

Read 2 more answers

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

4 years ago