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Aneli [31]
3 years ago
14

It’s kinda easy but I wanna make sure it’s right pls help if you can

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

a. 9/10

Step-by-step explanation:

hopes this helps

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The annual snowfall in city Ais 43.9 inches. This is 13.5 inches more than four times the snowfall in city B. Find the annual sn
Jlenok [28]

Answer:

24.475

Step-by-step explanation:

1st Divide: 43.9÷4=10.975

2nd Add 13.5: 10.975+13.5=24.475

--------------------------------------------------------------------

Hope this helps you out :)

7 0
3 years ago
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
The radius of a circle r in centimeters can be found by solving r to the second power = 70. What is the radius of the circle r,
slavikrds [6]

Answer:

bhb jkabdjkea 7- 70 bhakbhkd

3 0
3 years ago
I will give extra points please help
forsale [732]

Answer:

13

Step-by-step explanation:

b= 3

3x3= 9

9+4= 13

3x3+4=13

Hope this helps

7 0
3 years ago
A notebook cost $1.50 and a binder cost $6.50 jesica bought x notebooks and y binders
Bond [772]
1.50x + 6.50y is your algebraic expression.
4 0
3 years ago
Read 2 more answers
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