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3 years ago

Explain how the mean, median, and mode affect your purchase decisions?

Mathematics

1 answer:

Goryan [66]3 years ago

4 0

Answer:

When a frequency distribution is symmetric, the mean, median and mode coincide in their value (X = Me = Mo). ... In a distribution skewed to the right, the relationship is reversed, the mode is greater than the median, and this in turn is greater than the mean (Mo> Me>).

Step-by-step explanation:

l hope and this can help you

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Coefficient of (2x+y^2)^5

BigorU [14]

(2x + y2)5 heres you answer your so welcome

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3 years ago

What is the standard form of 0.00849

adelina 88 [10]

Answer:

8.49 × 10⁻³

Explanation:

Standard form of a number: A. × 10ᵇ

(where 'A' is the number, b is the exponent over 10)

For the number "0.00849", move the decimal point 3 places right so the exponent of 10 is -3.

Standard form: 8.49 × 10⁻³

Other examples of standard/scientific forms are: standard forms

i) 0.043, move the decimal point 2 places right = 4.3 × 10⁻²

ii) 4030, move the decimal point 3 places left = 4.03 × 10³

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1 year ago

Solve the equation: x3−20=−15

stich3 [128]

Answer:

x = 1.71

Step-by-step explanation:

$x^3-20=-15\\x^3=-15+20\\x^3=5\\\sqrt[3]{x^3} =\sqrt[3]{5} \\x=1.71$

7 0

2 years ago

Read 2 more answers

Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

$P[C]=p^n$

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.