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3 years ago

1)the difference between the upper and the lower quartile is the

1 answer:

3 0

A) interquartile range

The difference between the upper and the lower quartile is a measure of spread known as the interquartile range.

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The equation below represents Function A and the graph represents Function B:

kozerog [31]

The slope of function A is the coefficient of x, which is 6.

The slope of function B is the rise divided by the run, which is 3/1 = 3.

The appropriate choice is ...

... c) Slope of Function A = 2×Slope of Function B

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3 years ago

What is 2/9s of 630

maria [59]

Answer: its 18.27

Step-by-step explanation:

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3 years ago

Read 2 more answers

The augmented matrix of a consistent system of five equa- tions in seven unknowns has rank equal to three. How many parameters a

dangina [55]

Answer:

It is necessary four parameters.

Step-by-step explanation:

Since the augmented matrix of the consistent system has rank equal to three then the number of free variables is (number of unknows)-rank then there are 7-3=4 free variables. This means that are necessary four parameter to specify all solutions.

5 0

3 years ago

Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f

castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean $(\mu) = 115 \times 20$

= 2300

Standard deviation = $20 \times \sqrt{115}$

Standard deviation (SD) = 214.4761

TO find:

a) $P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{1200}{214.4761})$

$P(x > 3500)= P(Z >5.595)$

From the Z-table, since 5.595 is > 3.999

$P(x > 3500)=1-0.9999$

P(x > 3500) = 0.0001

Here, the replacement time for the mean $(\mu) = \dfrac{0+0.5}{2}$

= 0.25

Replacement time for the Standard deviation $\sigma = \dfrac{0.5-0}{\sqrt{12}}$

$\sigma = 0.1443$

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = $\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}$

= $\sqrt{(115\times 400) +(114\times 0.02082249}$

= $\sqrt{(46000) +2.37376386}$

= $\sqrt{(46000) +(2.37376386)}$

= $\sqrt{46002.374}$

= 214.482

Now; the required probability:

$P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})$

$P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})$

$P(x > 4125) = P(Z >8.376)$

$P(x > 4125) =1- P(Z$

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

7 0

2 years ago

Help on this math stuff

liq [111]

Plan A would be better

6 0

3 years ago