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tino4ka555 [31]
2 years ago
15

Need help pls. B is a boat on a bearing of 60degrees what is the nearing from A to B.

Mathematics
1 answer:
Verizon [17]2 years ago
3 0

Answer:

Step-by-step explanation:

the bearing of A from B is 065º. ... A boat sails from a

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Plz help ASAP I will add branliest to the first answer, I need to solve for the letter
harkovskaia [24]

Answer:

6. r=2

8.x=3

Step-by-step explanation:

6. 4r-3=5

4r-3=5 add 3 to both sides

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8. 5x-6=9

5x-6=9 add 6 to both sides

5x=15 divide 5 by both sides

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3 years ago
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Please help me with these!
makvit [3.9K]
No.4- This is not a triangle since one side should be less than the sum of the 2 other sides. In this case 8=3+5, hence it's equal & not smaller

N0.5- This is an isosceles triangle & there are an infinity of triangles with the mentioned degrees. (these are called similar triangle)

No6- This is a unique triangle

No.7- If the plane is parallel to the bases, then the slicing is a cercle.
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3 years ago
Magnitude of an earthquake M = log1
belka [17]
M = 4
Because the base of log is 10
10,000 is 10*4, and we know that log 10,000 =4, so M=4
6 0
3 years ago
Answer this math question and ill give you brainliest
natulia [17]

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3 years ago
Consider 8 blood donors chosen randomly from a population. The probability that the donor has type A blood is .40. Which of the
belka [17]

Answer: Option B is the only correct option.

Step-by-step explanation:

Number of samples = n = 8.

Probability of success = p = 0.4

Probability of failure = q = 0.6

r = chosen number of donors among the 8

To solve this question, we use the distribution formula

P(x=r) = nCr * p^r * q^n-r

For option A, to check if P(3<x<5) = 0.37. [3 and 5 inclusive]

When x = 3

P(x=3) = 8C3 * 0.4^3 * 0.6^5

P(x=3) = 56 * 0.064 * 0.07776

P(x=3) = 0.2787

When x= 4

P(x=4) = 8C4 * 0.4^4 * 0.6^4

P(x=4) = 70 * 0.0256 * 0.1296

P(x=4) = 0.2322

Since p(x=3) + p(x=4) is already greater than 0.37, then we know option A is NOT correct.

For option B, To check if the probability of 1 or fewer donor is about 0.11. i.e if P(x</=1) = 0.11

When x=o

P(x=0) = 8C0 * 0.4^0 * 0.6^8

P(x=0) = 1* 1 * 0.016796

P(x=0) = 0.016796.

When x = 1

P(x=1) = 8C1 * 0.4^1 * 0.6^7

P(x=1) = 8 * 0.4 * 0.02799

P(x=1) = 0.08958

P(x=0) + P(x=1) = 0.016796 + 0.08958

P(x=0) + P(x=1) = 0.10635.

Since this is approximately 0.11, then option B is a correct option.

For option C to check if the probability 7 or more donors not having type A = 0.0087

To do this,we determine thw probability of 7 or more donors having type A and we subtract our answer from 1.

First, we determine P(x>/=7)

When x= 7

P(x=7) = 8C7 * 0.4^7 * 0.6^1

P(x=7) = 8 * 0.001638 * 0.6

P(x=7) = 0.007864

When x=8

P(x=8) = 8C8 * 0.4^8 * 0.6^0

P(x=8) = 1 * 0.0006554 * 1

P(x=8) = 0.0006554

P(x=7) + P(x=8) = 0.007864 + 0.0006554 = 0.00852.

Since probability of 7 or more donors having type A is 0.00852 as against what was stated in the option C, then option C is NOT a correct option.

For option D, to check if the probability of exactly 5donors having type A blood = 0.28

When x=5

P(x=5) = 8C5 * 0.4^5 * 0.6^3

P(x=5) = 56 * 0.01024 * 0.216

P(x=5) = 0.1239.

Since probability of what was derived for having exactly 5 donors having sample A is different from what wqs given in the option, then option D is NOT correct.

For option E, since what was stated in the option negates what was derived for exactly 5 donors, then option E is NOT correct

6 0
3 years ago
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