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3 years ago

I75,986

Mathematics

2 answers:

velikii [3]3 years ago

8 0

Step-by-step explanation:

1a. 75,986

1b. Seventy five thousand nine hundred eighty six.

1c. 75,986=this one

$7 \times10000 + 5 \times 1000 + 9 \times 100 + 8 \times 10 + 6 \times 1$

1d. 70,000

1e 80

I did not understand the rest. Hope it's useful..... 

harina [27]3 years ago

4 0

Answer:

heres that plot you needed for that one glasses of water thing

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3x+-6+2 > = 5x -8......can someone please solve this

Maksim231197 [3]

Answer:

x ≤ 2

Step-by-step explanation:

If you referencing ">=" as greater than or equal to, follow the solution below:

Solve:

3x - 6 + 2 ≥ 5x - 8

Combine like terms.

3x - 4 ≥ 5x - 8

Subtract 5x from both sides.

-2x - 4 ≥ -8

Add 4 to both sides.

-2x ≥ -4

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3 years ago

Which expression is represented by the phrase "9 less than 5 times a number"?

sukhopar [10]

5x - 9.

Five times a number indicates multiplication and less than indicates subtraction. A number would be a variable since it isn’t specified.

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3 years ago

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pishuonlain [190]

Answer:

false

Step-by-step explanation:

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4 years ago

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Giving 20 points to the right person if your wrong your getting reported

Paha777 [63]

Answer:

730

Step-by-step explanation:

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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.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X 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