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AleksandrR [38]
3 years ago
11

If the Q1 is 77, and the Q3 is 93, what is the IQR (Inter Quartile Range) of the plot?

Mathematics
2 answers:
denis-greek [22]3 years ago
8 0

Answer:

The IQR (Inter Quartile Range) is 6.25

Step-by-step explanation:

Please give me brainlyist.

Murljashka [212]3 years ago
8 0
The IQR is 6.25 because it’s between numbers 77 and 93
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You can multiply the second equation by -2 

-5x + y = 15         everything by -2

10x - 2y = -30

now you can solve them

2x + 2y = -6
10x - 2y = -30

y cancels

2x= -6
10x = -30


12x = -36
/12.     /12

x = -3 


now to solve for y you substitute x for -3

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4 0
3 years ago
How to solve it and understanding it
Yanka [14]
All the values that go into a function

The output values are called the range.

Domain → Function → Range

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f(x)=y=x^2
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4 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.

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2 years ago
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Answer:

Complete y = Mx + b

Step-by-step explanation:

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