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nadya68 [22]
3 years ago
13

. Suppose that the average score in a statistics test is 84 with a standard deviation

Mathematics
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

96

Step-by-step explanation:

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Is d= -12 a solution to the equation 8d+6= -8d-186
kupik [55]

Answer:

It is a solution

Step-by-step explanation:

8d+6= -8d-186

Add 8d to each side

8d+8d+6= -8d+8d-186

16d +6 = -186

Subtract 6 from each side

16d +6-6 = -186 -6

16d = -192

Divide each side by 16

16d/16 = -192/16

d = -12

8 0
3 years ago
Which properties did Marty use?<br><br> Thank you very much!!
nadezda [96]

Answer:123

Step-by-step explanation:

5 0
3 years ago
Plz help ASAP thxs."..
Sliva [168]

Answer:

A

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A certain manufacturing plant produces electric fuses of which 20% are defective. Find the probability that in a sample of 8 fus
OverLord2011 [107]

Answer:

0.419

Step-by-step explanation:

To calculate this, we shall make use of Bernoulli approximation of Binomial distribution

if 20% are defective, then 80% are not defective

Probability of selecting a defective fuse is 20/100 = 0.2

Probability of selecting a non defective one is 0.8

Probability of at least 1 being defective is = 1 - Probability of none being defective

Mathematically that will be;

0.8^8 = 0.168

The above is probability of none defective

So the probability of at least one will be

1 - 0.168 = 0.832

Probability of not more than 1 means;

Probability of none + probability of 1 being defective

We already have probability of none above

Probability of 1 being defective means 8 will be non defective

The probability in this case is;

8 C 1 0.2^1 0.8^7

= 8 * 0.2 * 0.8^7 = 0.336

Add this to the probability of none = 0.336 + 0.168 = 0.504

So the probability that we want to calculate from the question will be;

Probability of at least one defective * probability of not more than one defective

= 0.832 * 0.504 = 0.419

6 0
3 years ago
Find the zeros of polynomial x^2-3
Yakvenalex [24]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The required zeroes are ~

\boxed{ \boxed {\sqrt{3}} \:  \:  and  \:  \: \boxed{- \sqrt{3} }}

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

Let's equate the given polynomial with 0, to find its roots ~

that is ~

  • {x}^{2}  - 3 = 0

  • {x}^{2}  - ( \sqrt{3} ) {}^{2}  = 0

  • (x +  \sqrt{3} )(x -  \sqrt{3} ) = 0

there are two cases,

Case # 1 - when (x + √3) = 0

  • x +  \sqrt{3}  = 0

  • x =  -  \sqrt{3}

and

Case # 2 - when (x - √3) = 0

  • x -  \sqrt{3}  = 0

  • x =  \sqrt{3}

4 0
2 years ago
Read 2 more answers
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