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2 years ago

14

You originally draw a design for an art contest on a 3 in. x 5 in. card. The second phase of the contest requires the drawing to

be transferred to an 8.5 in x 11 in. standard sheet of paper and utilize as much of the space on the paper as possible. You determine that the largest size one of the dimensions of your drawing can be is 10.5 in. What is the length of the other dimension if the two drawings are similar? Type your exact answer in the blank without the units, and round to the nearest tenths.

2 answers:

bagirrra123 [75]2 years ago

6 0

The answer is 6.3

3 / 5 = x / 10.5
x = 6.3

makvit [3.9K]2 years ago

4 0

Since the drawings are similar, the lengths of the 2 sides are proportional.

Since the largest size is 10.5 for the new drawing, it corresponds to the large side of the former drawing, which is 5.

Similary, the corresponding side of 3 (in the former drawing) in the larger drawing is what we are seeking to find, let's call this x.

So the ratio of 10.5 is to 5 should be equal to the ratio of x is to 3. Setting up the ratio, we have:

$\frac{10.5}{5}= \frac{x}{3}$

Now cross multiply and solve for x, we get:

$\frac{10.5}{5}= \frac{x}{3} \\3*10.5=5x\\31.5=5x\\\frac{31.5}{5} =x\\6.3=x$ .

So the length of the side we are looking for in the former drawing is 6.3 inches.

ANSWER: 6.3 in

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Step-by-step explanation:

In this problem I will assume that M is a point between point L and point N

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Answer:

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And replacing using the mass function we got:

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And adding the values we got:

$P(X\geq 20) = 0.8381$

Step-by-step explanation:

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

$X \sim Binom(n=25, p=0.85)$

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

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

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

We want to find the following probability:

$P(X\geq 20)= P(X=20)+P(X=21)+P(X=22)+P(X=23)+P(X=24)+P(X=25)$

And replacing using the mass function we got:

$P(X=20)=(25C20)(0.85)^{20} (1-0.85)^{25-20}=0.156$

$P(X=21)=(25C21)(0.85)^{21} (1-0.85)^{25-21}=0.211$

$P(X=22)=(25C22)(0.85)^{22} (1-0.85)^{25-22}=0.217$

$P(X=23)=(25C23)(0.85)^{23} (1-0.85)^{25-23}=0.161$

$P(X=24)=(25C24)(0.85)^{24} (1-0.85)^{25-24}=0.0759$

$P(X=25)=(25C25)(0.85)^{25} (1-0.85)^{25-25}=0.0172$

And adding the values we got:

$P(X\geq 20) = 0.8381$

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Read 2 more answers

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Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

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$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

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Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

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3 years ago