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zimovet [89]
3 years ago
13

Please help me ill give brainliest ​

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
6 0
The answer you are looking for is 9!
Brrunno [24]3 years ago
4 0

Answer:

9

-5 - 7(-2)

-5 - (-14)

+9

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Company B is selling at a better price.
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A sphere has a diameter of 22 units what is the radius and volume
lilavasa [31]

Answer:


Step-by-step explanation:

11

7 0
3 years ago
A polynomial function can be written as (x + 1)(x + 4)(x – 7). What are the x-intercepts of the graph of this function?
attashe74 [19]

Answer:

fourth option

Step-by-step explanation:

Given

f(x) = (x + 1)(x + 4)(x - 7)

To find the x- intercepts let f(x) = 0, that is

(x + 1)(x + 4)(x - 7) = 0

Equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

x + 4 = 0 ⇒ x = - 4

x - 7 = 0 ⇒ x = 7

x- intercepts are (- 1, 0 ), (- 4, 0 ), (7, 0 )

6 0
3 years ago
Find the value of x..........
likoan [24]

hey do you want to talk

4 0
2 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
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