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3 years ago

The length of a rectangle is 10 mm longer than its width. Its perimeter is more than 80 mm. Let w equal the

2 answers:

8 0

We know that the length (L) of the rectangle in question is 7mm longer than its width (W). Let's represent that as the following:
L=7+W

A rectangle's perimeter (the total sum of its sides) will be made my 2 sides representing the length (2L) and 2 sides representing the width (2W). We also know that this rectangle's perimeter is greater than 62. Since eventually we are solving for W, let's state all expressions in terms of W:
2L=2(7+W)
2(7+W)+2W>62
14+2W+2W>62
14+4W>62
4W>62-14
4W>48
W>48/4
W>12
If the rectangle's perimeter is greater than 62, then the width will be greater than 12. Let's confirm this:
Perimeter=2L+2W
P=2(7+12)+2(12)
P=14+24+24
P=62
<span>So we can see that if the perimeter is to surpass 62, W needs to be greater than 12 and L ( which is also 7+W) needs to be greater than 19.</span>

jok3333 [9.3K]3 years ago

3 0

P=L+L+W+W=2(L+W)
P>80
2(L+W)>80
divide 2
L+W>80

L is 10 more than W
L=10+W

A. L=10+W

B. L+W>80

C.
L+W>80
sub L=10+W
10+W+W>80
minus 10
2W>70
divid 2
W>35
W is mor than 35

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Answer:

$P(X \geq 3) = 1-P(X$

If we find the individual probabilities we gotL

$P(X=0)=(7C0)(0.51)^0 (1-0.51)^{7-0}=0.0068$

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And replacing we got:

$P(X \geq 3) = 1- [0.0068 +0.0494 +0.1543]= 0.7895$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=0.51)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

For this case we want to find this probability:

$P(X \geq 3)$

And we can use the complement rule for this case:

$P(X \geq 3) = 1-P(X$

If we find the individual probabilities we gotL

$P(X=0)=(7C0)(0.51)^0 (1-0.51)^{7-0}=0.0068$

$P(X=1)=(7C1)(0.51)^1 (1-0.51)^{7-1}=0.0494$

$P(X=2)=(7C2)(0.51)^2 (1-0.51)^{7-2}=0.1543$

And replacing we got:

$P(X \geq 3) = 1- [0.0068 +0.0494 +0.1543]= 0.7895$

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