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3 years ago

Help pleaseeeeeeee.

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer: 11x

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Explanation:

Let L be the length of rectangle B
There are two copies of L (along the top and bottom of the rectangle). The vertical pairs of sides are both 7x each.

For the triangle, we have three sides of 12x since this is an equilateral triangle. All three sides are congruent for any equilateral triangle.

The perimeter of the triangle is
P = s1+s2+s3
P = 12x+12x+12x
P = 36x

The perimeter of the rectangle is
P = 2*L+2*W
P = 2L+2*7x
P = 2L+14x

Since both perimeters are the same, this means
perimeter of triangle = perimeter of rectangle
36x = 2L+14x
36x-14x = 2L+14x-14x
22x = 2L
2L = 22x
2L/2 = 22x/2
L = 11x

So the length of the rectangle, in terms of x, is 11x. This is the final answer.

Note: if we knew the value of x, then we could find the numeric value of the length for the rectangle. But since we don't know x, we leave it as 11x.

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3 years ago

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(b) 0.6096

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We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) Two computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

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= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) Eight computers are chosen at random.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

LET X = Number of computers that are classified as ancient

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

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$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

LET X = Number of computers classified as cutting dash edge

So, it means X ~ $Binom(n=8, p=0.06)$

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P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

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