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4 years ago

Solve for h in h+6/5=2

Mathematics

1 answer:

Ivenika [448]4 years ago

6 0

Answer:

h=4/5

Step-by-step explanation:

$h + \frac{6}{5} = 2$

Collect like terms and simplify

$h = 2 - \frac{6}{5} \\ h = \frac{4}{5}$

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In the rolling of two fair dice calculate the following: P(Sum of the two dice is 7) = ______

vesna_86 [32]

Answer:

P(Sum of the two dice is 7) = 6/36

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

A fair dice can have any value between 1 and 6 with equal probability. There are two fair dices, so we have the following possible outcomes.

Possible outcomes

(first rolling, second rolling)

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)

(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)

(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)

(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)

(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

There are 36 possible outcomes.

Desired outcomes

Sum is 7, so

(1,6), (6,1), (5,2), (2,5), (3,4), (4,3).

There are 6 desired outcomes, that is, the number of outcomes in which the sum of the two dice is 7.

Answer

P(Sum of the two dice is 7) = 6/36

7 0

4 years ago

Question 5 of 10

Mkey [24]

Answer:

Your correct answer is C. 8

Step-by-step explanation:

I have taken this test just recently and I can assure you that it is the correct answer.

8 0

2 years ago

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Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab

Sphinxa [80]

Answer: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

$\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5$

Using the poisson process, we get that

(a) less than or equal to 12

probability= $P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980$

(b) equal to 13

probability= $P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013$

probability= $P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020$

(d) equal to 20

probability= $P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026$

(e) between 10 and 15, inclusively

probability= $P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318$

Hence, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0

3 years ago

F+4/g=6 Solve for F

Svet_ta [14]

$\dfrac{f+4}{g}=6\ \ \ \ |multiply\ both\ sides\ by\ g\\\\f+4=6g\ \ \ \ |subtract\ 4\ from\ both\ sides\\\\\boxed{f=6g-4}$

5 0

3 years ago

Determine if the triangle is Right, Acute or Obtuse.

Lyrx [107]

Answer:

I think the right answer is: Acute

8 0

3 years ago

Read 2 more answers