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trasher [3.6K]
2 years ago
14

(see image) please help I need help with this math problem * i give brainlist

Mathematics
1 answer:
Leni [432]2 years ago
8 0
Answer: a) g(x)=(x-5)^2-6
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Evaluate: 16 - 2+3 - 4
natulia [17]

Answer: 13

Step-by-step explanation:

Following order of operations:

16-2+3-4

14+3-4

17-4

13

8 0
3 years ago
Read 2 more answers
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
Thomas Vega offers to pay $59 of the March cell phone bill. Each of the other 4 members of the family agrees to split the rest o
bija089 [108]
59×2=118 cost of bill.
59÷4=14.75 for the other 4 if the bill was 118 assuming Thomas pays half.

or...59 ×5=295
assuming they all pd 1/5 of the bill
6 0
3 years ago
What is the the volume of this?<br><br> Also you can explain just make the explanation short
fgiga [73]
15×20×10=3000
10×20×15=3000
3000+3000=6000 cm^{2}

6 0
3 years ago
Solve the absolute value equation |8|=28
DENIUS [597]

The answer is false. This is because<u> 8 does not equal the number 28</u>.

Explination:

    |8| = 28

     8≠28

7 0
3 years ago
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