Can someone please help me

Mathematics

1 answer:

agasfer [191]3 years ago

6 0

Answer:

See below

Step-by-step explanation:

If |x| < |y| and both are negative the x will be to the right of y on the number line. The point x is loser to the zero on the number line. We can illustrate this by supposing x = -3 and y = -6.

|x| = 3 and |y| = 6 and 3 < 6 but -3 on the number line is higher ( to the right) of -6 and closer to the zero.

|-8 1/3| = 8 1/3

|7 2/3| = 7 2/3

The absolute values of 8 1/3 > than absolute value of 7 2/3. The point 7 2/3 is closer to the zero on a number line.

The absolute value of F is greater than absolute value of E so E is closer to the zero line than F.

You might be interested in

Help asap please!!!! due tonight!!! brainliest for fastest right answer!!! xx

Serga [27]

Answer:

Step-by-step explanation:

3 0

1 year ago

fter production, a computer component is given a quality score of A, B, and C. On the average U of the components were given a q

RoseWind [281]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given question some of the data is missing so, its correct question is defined in the attached file please find it.

Let

A is quality score of A

B is quality score of B

C is quality score of C

$\to P[A] =0.55\\\\\to P[B] =0.28\\\\\to P[C] =0.17\\$

Let F is a value of the content so, the value is:

$\to P[\frac{F}{A}] =0.15\\\\\to P[\frac{F}{B}] =0.12\\\\\to P[\frac{F}{C}] =0.14\\$

Now, we calculate the tooling value:

$\to p[\frac{C}{F}]$

using the baues therom:

$\to p[\frac{C}{F}] = \frac{p[C] \times p[\frac{F}{C} ]}{p[A] \times p[\frac{F}{A}] + p[B] \times p[\frac{F}{B}]+p[C] \times p[\frac{F}{C}] }$

$= \frac{ 0.17 \times 0.14 }{0.55 \times 0.15 + 0.28 \times 0.12 + 0.17 \times 0.14 } \\\\= \frac{0.0238}{0.0825 + 0.0336 + 0.0238} \\\\= \frac{0.0238}{0.1399} \\\\=0.1701$