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2 years ago

6(1 − 3d) < 2(7d + 6

Mathematics

2 answers:

vagabundo [1.1K]2 years ago

8 0

Answer:

d> -3/16

Step-by-step explanation:

Andre45 [30]2 years ago

8 0

Answer:

\begin{bmatrix}\mathrm{Solution:}\:&\:d>-\frac{3}{16}\:\\ \:\mathrm{Decimal:}&\:d>-0.1875\\ \:\mathrm{Interval\:Notation:}&\:\left(-\frac{3}{16},\:\infty \:\right)\end{bmatrix}

Step-by-step explanation:

6-18d-6<14d+12-6

-18d<14d+6

-32d<6

\frac{32d}{32}>\frac{-6}{32}

d>-\frac{3}{16}

Brainliest plz

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16. From past experience, a company has found that in cartons of transistors, 92% contain no defective transistors, 3% contain o

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Step-by-step explanation:

The mean of a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)+...+x_n*p(n)$

where $x_1,x_2,...,x_n$ are the values that the variable can take and $p(x_1),p(x_2),...,p(x_n)$ are their respective probabilities.

So, if we call x the number of defective transistors in cartons, we can calculate the mean E(x) as:

$E(x)=(0*0.92)+(1*0.03)+(2*0.03)+(3*0.02)=0.15$

Because there are 0 defective transistor with a probability of 0.92, 1 defective transistor with a probability of 0.03, 2 defective transistors with a probability of 0.03 and 3 defective transistors with a probability of 0.01.

At the same way, the variance V(x) is calculated as:

$V(x)=E(x^2)-(E(x))^2$

Where $E(x^2)=x_1^2*p(x_1)+x_2^2*p(x_2)+...+x_n^2*p(n)$

So, the variance V(x) is equal to:

$E(x^2)=(0^2*0.92)+(1^2*0.03)+(2^2*0.03)+(3^2*0.02)=0.33\\V(x)=0.33-(0.15)^2\\V(x)=0.3075$

Finally, the standard deviation is calculated as:

$S(x)=\sqrt{V(x)} \\S(x)=\sqrt{0.3075} \\S(x)=0.5545$

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