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miss Akunina [59]
3 years ago
7

What is 5,683 divided by 8

Mathematics
1 answer:
salantis [7]3 years ago
7 0

Answer:710.375‬

Step-by-step explanation:

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omeli [17]
4/15 because you make it into a fraction: 8/30, you find the GCF which is 2, so then you divide both the numerator and denominator by 2 and get 4/15 at it's simplest term
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Can you help me solve for z<br><br><br> Please shower work so I can understand
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z=(x-1)/2

Step-by-step explanation:

this is the correct answer

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4 years ago
What is 84(0.5) not in decimal form
SpyIntel [72]

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42

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6 0
3 years ago
Read 2 more answers
PLEASE HURRY-When Melissa is playing soccer she makes 2 goals out of every 12 shots on goal. How many times would you expect her
melisa1 [442]

Answer: 24 goals

Step-by-step explanation:

Melissa makes 2 goals out of every 12 shots on goal.

This means that the amount of shots needed to score a goal for her is:

= Number of shots / Number of goals

= 12 / 2

= 6 shots

She scores a goal every 6 shots.

If she makes 144 shots therefore, she will score:

= Number of shots / Number of shots needed per goal

= 144 / 6

= 24 goals

6 0
3 years ago
Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,
OLEGan [10]
Let D denote the event that an HD is defective, and F_i the event that a particular HD was produced at facility i.

You are asked to compute

\mathbb P(F_2\mid D)
\mathbb P(F_4\mid D)
\mathbb P(D^C)

From the definition of conditional probabilities, the first two will require that you first find \mathbb P(D). Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so \mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015.

Use the law of total probability to determine the value of the denominator:

\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

\mathbb P(D)=0.015+0.02+0.01+0.03=0.075

and thus

\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04

(b) Similarly,
\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16

(c)
\mathbb P(D^C)=1-\mathbb P(D)=0.925
6 0
4 years ago
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