The answer appears in the image, this is for practice and please tell me an explanation on how you got the answer.

Please answer all with explanation

stepladder [879]

3b.4+y/3=2

12+y/3=2

12+y=6

y=6-12

y=-6

.

8 0

3 years ago

I’m not to sure on how to find CEB can someone explain how to find it and what the answer would be,please

lakkis [162]

Answer:

m∠CEB is 55°

Step-by-step explanation:

Since ∠ADE = 55°, and ∠ADE is half of ∠ADC because ED bisects ∠ADC. Bisect means to cut in half.

∠ADC = 110° because it is double of ∠ADE.

Since AB║CD and AD║BC, the two sets of parallel lines means this shape is a parallelogram. In parallelograms, opposite angles have equal measures.

∠ADC = ∠CBE = 110°

All quadrilaterals have a sum of angles 360°. Since ∠DCB = ∠BAD and we know two of these other angles are each 110°:

360° - 2(110°) = 2(∠DCB)

∠DCB = 140°/2

∠DCB = ∠BAD = 70°

∠DCB was bisected by EC, which makes each divided part half.

∠DCE = ∠BCE = (1/2)(∠DCB)

∠DCE = ∠BCE = (1/2)(70°)

∠DCE = ∠BCE = 35°

All triangles' angles sum to 180°.

In ΔBCE, ∠BCE = 35° and ∠CBE = 110°.

∠CEB = 180° - (∠BCE + ∠CBE)

∠CEB = 180° - (35° + 110°)

∠CEB = 55°

Therefore m∠CEB is 55°.

5 0

3 years ago

The ratio of men and women at a class is 6 to 5. how many women are their if their are 3600 men?

umka2103 [35]

Answer:

3,000

Step-by-step explanation:

6:5

1.2( divided 6 by 5):1 (Divide 5 by 5)

3600 divided by 1.2

3,000

4 0

3 years ago

If five quarts of paint are needed for 250 square-feet of room, 19 quarts of paint will cover how many square feet of room?

Allushta [10]

Answer:

check the attachments for the solution

6 0

2 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