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3 years ago

Help me please if you 100% know it:)

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

5 0

Its 4 people because there are 2 who studied 4 hours, 1 who studied 5, and another one for 8.5, and 2 + 1 + 1 = 4

vichka [17]3 years ago

4 0

The answer is 4 because 2 studied for 4 hours and 1 studied for 5 hours and 1 studied for 8 hours. the number of students that studied AT LEAST 4 hours, so this would be described by the inequality x>=4

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A sphere with a diameter of 30 inches is placed in a cube. The sphere touches the lateral faces

Nina [5.8K]

Answer:

a) 30 inches × 30 inches × 30 inches

b) 27,000 inches³

c) 14,130 inches³

d) 12,870 inches³

Step-by-step explanation:

Edge of the cube = diameter = 30 inches

Volume of cube = s³

30³ = 27000 inches³

Radius = 30/2 = 15

Volume of sphere:

(4/3) × 3.15 × 15³

14130 inches³

Difference:

27000 - 14130

12870 inches³

5 0

3 years ago

Me ayudan xfaaaaaa (':

IceJOKER [234]

Answer:

$2(z + 8) 3 \\$

4 0

3 years ago

Suppose that commercial airplane crashes in a certain country occur at the rate of 2.5 per year. what is the probability that fo

lesya [120]

The poisson distribution applies here.
$\lambda=2.5$
$P(X=4)=\frac{e^{-2.5}\times2.5^{4}}{4!}=0.134$

4 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

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

Solve 3x − x + 2 = 12

marissa [1.9K]

3x − x + 2 = 12
2x + 2 = 12
2x = 10
x = 5

answer
x = 5

5 0

3 years ago

Read 2 more answers