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2 years ago

Make a conjecture abouth the following statement: Angle ABC and angle CBD are adjacent.

Mathematics

1 answer:

zvonat [6]2 years ago

3 0

The measure of Angle ABD = measure of Angle ABC + measure of Angle CBD

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Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5i a. 29i b. 291-20 C.-21i d.29

attashe74 [19]

Answer:

A. 29i

Step-by-step explanation:

Step 1: Plug in given variables

(2 - 5i)(2 + 5i)i

Step 2: Difference of squares (expand)

(4 - 25i²)i

Step 3: Imaginary numbers rules

(4 - 25(-1))i

Step 4: Combine like terms

(4 + 25)i

(29)i

Your final answer will be 29i

8 0

3 years ago

Read 2 more answers

3 times 2 cubed + 5 squared-18 divided by 6

Strike441 [17]

Answer:

Step-by-step explanation:

3 x 2^3 + 5^2 - 18/6

= 3 x 8 + 25 - 3

= 24 + 25 - 3

= 49 - 3

= 46

7 0

2 years ago

Lily had 95 inches of yarn. After cutting 4 pieces that were each the same length, she had 15 inches of yarn left.

Anika [276]

Each was 10 inches because she had 95 and 95-15=40 divided by 4 = 10

6 0

3 years ago

Read 2 more answers

Plz help with this problem pic down below

kherson [118]

Answer:

yes is is

Step-by-step explanation:

if you connect the lines it forms a straight line

3 0

3 years ago

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th

umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable X is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, p = 0.09.

An item is defective irrespective of the others.

The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

$=3E(X^{2})+E(X)+2$

Compute the expected value of X as follows:

$E(X)=np$

$=4\times 0.09\\=0.36$

The expected value of X is 0.36.

Compute the variance of X as follows:

$V(X)=np(1-p)$

$=4\times 0.09\times 0.91\\=0.3276\\$

The variance of X is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

$=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

$=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.

4 0

3 years ago