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3 years ago

Triangles ABC and DEF are similar find d

Mathematics

2 answers:

gavmur [86]3 years ago

6 0

Answer:

A=D where is picture . it is required

strojnjashka [21]3 years ago

6 0

Answer:

Step-by-step explanation:

i took the test

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Use the function rule y=5x+1. What is y when x=3

Dmitry_Shevchenko [17]

Answer:

y = 16

(3, 16)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

y = 5x + 1

x = 3

Step 2: Evaluate

Substitute in x: y = 5(3) + 1
Multiply: y = 15 + 1
Add: y = 16

7 0

3 years ago

Read 2 more answers

Plese help for brainlist !!!

Ipatiy [6.2K]

Answer:

{2, 4, 6}

Step-by-step explanation:

the domain is the interval or set of valid x (input) values.

the range is the interval or set of valid y (result) values.

all we need to do is use the given x values in the functional definition and collect the result values. that is the range.

x = -1

y = 2×-1 + 4 = 2

x = 0

y = 2×0 + 4 = 4

x = 1

y = 2×1 + 4 = 6

7 0

2 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

Find the volume. Image is below

DaniilM [7]

3,075ft^3 (cubic feet)

(the ^3 means to the power of 3. aka cubed)

3 0

3 years ago

2. How do you multiply and divide rational expressions?

tatuchka [14]

Answer:

1.Factor both the numerator and the denominator.

2.Write as one fraction.

3: Simplify the rational expression.

4: Multiply any remaining factors in the numerator and/or denominator.

Step-by-step explanation:

7 0

3 years ago