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2 years ago

Rewrite as a simplified fraction.

Mathematics

1 answer:

n200080 [17]2 years ago

4 0

16/5 is roughly equal to 3.2. Is that what your are asking?

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Need help with these two

DiKsa [7]

Am sorry I really need this point sorry again

4 0

3 years ago

Number 1 has a picture

MA_775_DIABLO [31]

Question 1

cos(C) = adj/hyp
cos(C) = BC/AC
cos(C) = 15/17

Answer: Choice C) 15/17

=============================================

Question 2

sin(E) = opp/hyp
sin(E) = FD/DE
sin(E) = 24/25

So angle E is the answer

=============================================

Question 3

AC is the adjacent side to the given angle A = 33 degrees
AB = 14 is the hypotenuse

cos(angle) = adjacent/hypotenuse
cos(33) = x/14
14*cos(33) = x
x = 14*cos(33)
x = 11.741
x = 11.7

Answer: 11.7

=============================================

Question 4

x = length of board

cos(angle) = adj/hyp
cos(55) = 6/x
x*cos(55) = 6
x = 6/cos(55)
x = 10.46068
x = 10.5

Answer: 10.5

=============================================

Question 5

112 is the adjacent side to the given angle 62 degrees
x = height of building, which is the opposite leg

tan(angle) = opp/adj
tan(62) = x/112
112*tan(62) = x
x = 112*tan(62)
x = 210.641364
x = 210.6

Answer: 210.6

3 0

3 years ago

Solve r = (9–9) for p.

Dmitry_Shevchenko [17]

Answer: u mean r=0

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

10 points if you solve this one problem.

zalisa [80]

Answer:

150

Step-by-step explanation:

The exterior angle is equal to the sum of the opposite interior angles

<1 + <2 = <3

3x+11 + 3x+19 = 7x+10

Combine like terms

6x+30 = 7x+10

Subtract 6x from each side

6x -6x+30 = 7x-6x+10

30 = x+10

Subtract 10 from each side

30-10 = x+10-10

20 =x

We need to find angle 3

<3 = 7x+10

= 7(20)+10

= 140+10

= 150

7 0

2 years ago

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

5 0

2 years ago