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3 years ago

The hull speed of a boat is approximated by the

Mathematics

2 answers:

sweet [91]3 years ago

7 0

Answer:

a = ✔ 1.34

b = ln ✔ – 10

ln > 10

Law Incorporation [45]3 years ago

5 0

a=1.34

b=ln-10

ln>10

Step-by-step explanation:

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PLEASE HELP ME I DONT UNDERSTAND

Vlad1618 [11]

Answer:

Picture #1: x=16 ; y=10

Picture #2: x=21 ; y=15

Picture #3: x=18

Step-by-step explanation:

Picture #1:

(8x-14)* and (5x+34)* are Alternate Exterior Angle so they are congruent. And (8x-14)* and (5y+16)* are supplementary so they equal 180*.

Picture #2:

(6x+7)* and (3x-16)* are supplementary so they equal 180*. And (6x+7)* and (11y-32)* are Vertical Angles so they are congruent.

Picture #3:

(7x-6)* and (4x-7)* are supplementary so they equal 180*.

6 0

3 years ago

An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side true or false

sashaice [31]

The answer is true. An altitude of a triangle is the perpendicular segment from a vertex to the line containing the opposite side.

3 0

2 years ago

What is 3 1/4 - 2 1/2

sergiy2304 [10]

The answer is just do easy subtraction 3/4

6 0

2 years ago

PLEASE HELP SMART PEOPLE WILL MARK BRAINLIEST

sergey [27]

Answer:

$\angle YPR = 42^o$

Step-by-step explanation:

Its given:

$\angle FPY = 90^o \\ \angle YPR + \angle FPR = 90^0 \\ 4p + 80 + 5p + 82 = 90^o \\ 9p = -72^o \\ p = -8 \\ \\ \angle YPR = 5(-8) + 82 = 42^o$

5 0

3 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

2 years ago