Next question, need help ASAP please

In a recent year, 304 of the approximately 300,000,000 people in the United States were struck by lightning. Estimate the prob

umka2103 [35]

Answer:

$\frac{19}{18,750,000}=0.0000010133333333$

Step-by-step explanation:

We have been given that in a recent year, 304 of the approximately 300,000,000 people in the United States were struck by lightning.

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Number of possible outcomes}}$

$\text{P( A randomly selected person in the United States will be struck by lightning this year)}=\frac{304}{300,000,000}$

$\text{P( A randomly selected person in the United States will be struck by lightning this year)}=\frac{19}{18,750,000}$

$\text{P( A randomly selected person in the United States will be struck by lightning this year)}=0.0000010133333333$

Therefore, the probability that a randomly selected person in the United States will be struck by lightning this year is $\frac{19}{18,750,000}=0.0000010133333333$ .

6 0

4 years ago

Read 2 more answers

IT COST $9 TO GO TO PETE'S POTTERY PLACE TO MAKE YOUR OWN BOWLS $3 PER BOWL. NATALIE GOES TO PETE'S POTTERY PLACE AND MAKES B BO

S_A_V [24]

C= 5(9*3b)
C= 5(9*12)
C= 5(108)
C=540

8 0

4 years ago

to increase the mean of 4 numbers by 2, by how much would the sum of the 4 numbers have to increase? check image.

solong [7]

Answer:

8

Step-by-step explanation:

x=average

(x+x+x+x)/4=average

to increase x by 2 you add 2

x+2=average

(x+2+x+2+x+2+x+2)=4=(4x+8)/4=x+2

you'd have to increase the sum by 8 to make the numbers' averages' go up by 2

8 0

3 years ago

Use the formula A=bhto find the area of the parallelogram. 2.1 m 6.3 m

Feliz [49]

Answer:

A = 13.23 m²

Step-by-step explanation:

A = bh

Where,

A = area of the parallelogram

b = base of the parallelogram

h = height of the parallelogram

From the question, assume

b = 2.1 m

h = 6.3 m

A = bh

= 2.1 m × 6.3 m

= 13.23 m²

A = 13.23 m²

7 0

3 years ago

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

(c) 0.906

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

3 years ago

Next question, need help ASAP please ​

Next question, need help ASAP please