All categories

3 years ago

What is 5.6 as a fraction or mixed number in simplest form

Mathematics

2 answers:

kherson [118]3 years ago

4 0

It would have to be 5 60/100

(5 wholes and sixty out of a hundred)

Alinara [238K]3 years ago

3 0

The answer, in simplest form, is 3/5.

You might be interested in

Graph g(x) = –2x – 3

Luba_88 [7]

Answer:

The slope would be -2 with a y-intercept of (0, -3).

I hope this helped to answer your question.

4 0

3 years ago

Independent Practice

snow_tiger [21]

Answer:

3/2; −3

Step-by-step explanation:

To find the x-intercept, substitute in 0 for y and solve for x . To find the y-intercept, substitute in 0 for x and solve for y .

x-intercept(s): ( 3 /2 , 0 )

y-intercept(s): ( 0 , − 3 )

8 0

3 years ago

Whats the m.a.d. of 0, 3, 6, 9, 11, 13, 14? n=7

KatRina [158]

Answer:

4 2/7

Step-by-step explanation:

It is convenient to let a spreadsheet or graphing calculator do the tedious math for you. (See the attachment.)

_____

The sum of the numbers is 56, so their average is 56/7 = 8. Then the absolute deviations from average are ...

... 8, 5, 2, 1, 3, 5, 6

and the total of these is 30. The mean absolute deviation is this sum divided by the number of numbers, 7, so is ...

... 30/7 = 4 2/7

_____

<em>More about Absolute Deviation from Average</em>

"Absolute" here refers to the absolute value function, which drops the sign of a negative number. The absolute deviation of the data value 0 is ...

... |0 - 8| = |-8| = 8

For 3, it is ...

... |3 - 8| = |-5| = 5

4 0

3 years ago

What's the answer to n+4.5-0.3n-3

VLD [36.1K]

1. Collect like terms

(n - 0.3n) + (4.5 - 3)

2. Simplify

0.7n + 1.5

Final Answer:

0.7n + 1.5

7 0

3 years ago

At an airport, 79% of recent flights have arrived on time. A sample of 7flights is studied. a.Compute the mean of this probabili

xxTIMURxx [149]

Answer:

a. 5.53

b. 1.078

c. 0.126

d. 0.109

e. 0.549

f. 0.834

g. 0.451

Step-by-step explanation:

The percentage of the flights that arrive on time, P(x) = 79%

The number of flights in the sample, n = 7 flights

a. The mean of the probability distribution, μ = ∑x·P(x)

Therefore, we have; μₓ = n·p

μₓ = 7 × 79/100 = 5.53

b. The standard deviation, σₓ = √(n·p·(1 - p))

∴ σₓ = √(7 × 0.79 × (1 - 0.79)) ≈ 1.078

c. We have;

p = 0.79

q = 1 - p = 1 - 0.79 = 0.21

By binomial probability distribution formula, we have;

The probability of exactly four, P(Exactly 4) = ₇C₄·p⁴·q³

P(Exactly 4) = 35 × 0.79⁴×0.21³ ≈ 0.12625

d. The probability of less than 4 is given as follows;

P(Less than 4) = ₇C₀·p⁰·q⁷ + ₇C₁·p¹·q⁶ + ₇C₂·p²·q⁵ + ₇C₃·p³·q⁴

∴ P(Less than 4) = 1×0.79^0 * 0.21^7 + 7 * 0.79^1 × 0.21^6 + 21*0.79^2*0.29^5+ 85×0.79^3*0.21^4 ≈ 0.109

The probability of less than 4 is ≈ 0.109

e. The probability that more than 5 is given as follows;

P(More than 5) = ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.549

f. The probability that at least 5 of the flight were on time is given as follows;

P(At least 5) = ₇C₅·p⁵·q² + ₇C₆·p⁶·q¹ + ₇C₇·p⁷·q⁰

∴ P(At least 5) = 21×0.79^5 * 0.21^2 + 7×0.79^6 * 0.21 + 1 * 0.79^7 × 0.21^0 ≈ 0.834

g. For the probability that no more than 5 of the flights were on time, e have;

P(At most 5) = 1 - P(More than 5)

∴ P(At most 5) = 1 - 0.549 ≈ 0.451.

6 0

2 years ago