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evablogger [386]
3 years ago
10

Help help i will give brainliest please look at picture

Mathematics
2 answers:
Goryan [66]3 years ago
4 0

Answer:

Your answer is right, it is 343cm

Step-by-step explanation:

14 \times 21  = 294 \\ 28 - 21 = 7 \\ (7 \times 14) \div 2 = 49 \\ 294 + 49 = 343

nydimaria [60]3 years ago
3 0

Answer:

B

Step-by-step explanation:

not completely sure that's right sorry if its Wong

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Which of these is an example of the Commutative property of addition? And Explain why? 3+5=4+4 or 3+5=5+3
arlik [135]

Answer:

3+5=5+3

Step-by-step explanation:

a+b=b+a

comunitive property

3 0
3 years ago
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Find the y-intercept of the following equation. Simplify your answer.<br> y = -10x -3/7
77julia77 [94]

Answer: (0, -3/7)

Step-by-step explanation:

The Y-intercept would be the value of y when x is at 0, which is when the Y axis is intercepted. -3/7 is the starting position of the function when the the X = 0.

7 0
3 years ago
Find the value of x
romanna [79]

they are 2 similar triangles, one is smaller than the other. they are in the ratio of 2:1,

hence 7x =56÷2

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x=4

4 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
Please help me. if you can help me answer multiple questions only 3. offering 30 points for it
lubasha [3.4K]

Answer:

(-9, -5)

Step-by-step explanation:

Ok, so when you move an image to the right, you are moving along the x-axis, and when you move up, you are moving up the y-axis. So if the altered image is (x,y) and the values are (-5, -1), you reverse what has been done to the image. In this case, since we moved to the right 4 units, we know that means we added 4 to x, so we subtract 4 to get -9. And then, for the y-value, because we added 4, we do the opposite, and subtract 4 to get -5. So the pre-image should be (-9, -5)

7 0
3 years ago
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