What is the measure of ∠DBE?

Determine the distance between (-5,2) and (-2,5). show your work. round to the nearest tenth if necessary.

statuscvo [17]

Answer

-2

Step-by-step explanation:

learn how to do your work f^gg0+

4 0

3 years ago

) Si on multiplie les deux membres d'une inégalité par un même nombre négatif non nul, on obtient une inégalité de même sens.

iogann1982 [59]

Answer:

John has 9 different flags that he wants to hang on the wall. How many different ways can the flags be arranged in a row?

Step-by-step explanation:

John has 9 different flags that he wants to hang on the wall. How many different ways can the flags be arranged in a row?

7 0

4 years ago

Which facts could be applied to simplify this expression? Check all that apply. (-5x-3y)-(-x-3z).

Mademuasel [1]

Answer:

The correct options are 1, 2, 4 and 6.

Step-by-step explanation:

The given expression is

$(-5x-3y)-(-x-3z)$

To subtract like term subtract like terms, subtract the coefficients, not the variables.

Therefore option 1 is correct.

Like terms are terms that contain the same variable, raised to the same power. For example: 3xy and 5xy, 4x² and 3x².

Therefore, option 2 is correct.

Using distributive property we get

$-5x-3y-(-x)-(-3z)$

$-5x-3y+x+3z$

Only combine terms which contain the same variable.

Therefore option 4 is correct.

$(-5x+x)-3y+3z$

$-4x-3y+3z$

The simplified expression is -4x-3y+3z. Therefore option 6 is correct.

3 0

3 years ago

Read 2 more answers

Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or

lana66690 [7]

Answer:

(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010

(b)

(1 – p) = 0.010

(c)

E(x) = 25000 x 0.010

= 259.2

Explanation has given below.

Step-by-step explanation:

Solution:

(a) Probability that a triplet is decoded.

2 out of three

P = 0.94, n = 3

m= no of correct bits

m bit (3, 0.94)

At p(m≤1) = B (1; 3, 0.94)

= 0.010

(b) Using your answer to part (a),

(1 – p) = 0.010

Error for 1 bit transmission error.

(c) How does your answer to part (a) change if each bit is repeated five times (instead of three?

P( m ≤ 2 )

L = Bit (5, 0.94)

= B (2; 5, 0.94)

= 0.002

(d) Imagine a 25 kilobit message (i.e., one requiring 25,000 bits to send). What is the expected number of errors if there is no bit repetition implemented? If each bit is repeated three times?

Given:

h = 25000

Bits are switched during transmission between two computers = 6% = 0.06

m = Bit (25000, 0.06)

E(m) = np

= 25000 x 0.06

= 1500

m = Bit (25000, 0.01)

E(m) = 25000 x 0.010

= 259.2

3 0

3 years ago

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has

Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e. $P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e. $P(\bar{X_2}|Y)=0.7$

So, $P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )= $1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )= $1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b) $P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)= $P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

D)

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$

3 0

3 years ago