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astraxan [27]
3 years ago
14

Which statement is true about the difference 2√7-√28

Mathematics
1 answer:
maw [93]3 years ago
7 0
It seems like there are answer choices but you forgot to post them. If you want to simplify then

2*sqrt(7) - sqrt(28) = 2*sqrt(7) - sqrt(4*7)
2*sqrt(7) - sqrt(28) = 2*sqrt(7) - sqrt(4)*sqrt(7)
2*sqrt(7) - sqrt(28) = 2*sqrt(7) - 2*sqrt(7)
2*sqrt(7) - sqrt(28) = (2-2)*sqrt(7)
2*sqrt(7) - sqrt(28) = (0)*sqrt(7)
2*sqrt(7) - sqrt(28) = 0

So the whole complicated expression simplifies to 0 (zero)

This is because sqrt(28) = 2*sqrt(7) when you simplify sqrt(28)

Side Note: "sqrt" is shorthand for "square root"

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Rashad draws a figure and divides it into equal parts. Two of the parta are red. The other 4 parts are blue. Rashad saya that 2/
Juliette [100K]
Let
x--------------> the complete figure 
we know that
the figure is divided into 6 equal parts
x= (x/6)+(x/6)+(x/6)+(x/6)+(x/6)+(x/6)
red=(x/6)+(x/6)--------> red=(2x/6)=x/3
blue=(x/6)+(x/6)+(x/6)+(x/6)=(4x/6)=2x/3

Rashad say that 2/4 of the figure is red-------> Rashad is wrong
because (2/4) <span>It is half of the figure

</span>the correct thing is to say (1/3) o the figure is red and (2/3) of the figure is blue
3 0
3 years ago
Lengths of newborn girls are normally distributed with a mean of 49.2 cm and a standard deviation of 1.8 centimeters. Consider a
Elza [17]
The answer is <span>1683.

Hope this helps.</span>
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3 years ago
For f(x) = 0.02(2)^x find the average rate of change from x= 3 to x= 8
Fudgin [204]

None of the offered choices is correct.

f(3) = 0.02·2³ = 0.02·8 = 0.16

f(8) = 0.02·2⁸ = 0.02·256 = 5.12

Then the average rate of change is

... (f(8) - f(3))/(8 - 3) = (5.12 -0.16)/5 = 4.96/5 = 0.992

5 0
3 years ago
A grocer mixed walnuts worth $1.02 per pound with peanuts worth $0.99 per pound. If she made 69 pounds of a mixture worth $1 per
Nataly [62]
The cost of the mixture is $69 *$1 = $69. If all 69 pounds were peanuts, the cost would be 69*$0.99 = $68.31, which is $0.69 less. Each pound of walnuts used instead of peanuts adds $0.03 to the cost of the mixture, so there were $0.69/$0.03 = 23 pounds of walnuts. 
<span>_____ </span>
<span>Let w represent the number of pounds of walnuts in the mixture. Then 69-w is the number of pounds of peanuts. The cost of the mixture will be </span>
<span>1.02w + 0.99(69-w) = 1.00*69 dollars </span>
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6 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
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