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3 years ago

The aspect ratio (the ratio of screen width to

Mathematics

1 answer:

Strike441 [17]3 years ago

5 0

Answer:

$Width = 36in$

$Size = 42.02$

Step-by-step explanation:

Given

$x \to width$

$y \to height$

$x : y = 16 : 9$

Required

The size of $screen$ of height $20,6in$

First, we calculate the width using the following equivalent ratios

$16: 9 = x : 20.6$

Express as fraction

$\frac{16}{ 9} = \frac{x }{ 20.6}$

Solve for x

$x = 20.6 * \frac{16}{ 9}$

$x = 36.62$

Hence:

$Width = 36in$ --- approximated

So, we have:

$x = 36.62$

$y =20.6$

The size (diagonal) is then calculated using:

$Size = \sqrt{x^2 + y^2$

$Size = \sqrt{36.62^2 + 20.6^2$

$Size = \sqrt{1765.3844$

$Size = 42.02$

$Size =42in$ --- approximated

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High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high

KengaRu [80]

Answer:

1. It is not appropriate to use the normal curve, since np = 7.4 < 10.

2. The probability that more than 32% of the people in this sample have high blood pressure is 0.0033 = 0.33%.

Step-by-step explanation:

Binomial approximation to the normal:

The binomial approximation to the normal can be used if:

np >= 10 and n(1-p) >= 10

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The proportion of U.S. adults with high blood pressure is 0.2. A sample of 37 U.S. adults is chosen.

This means, respectively, that $p = 0.2, n = 37$

Is it appropriate to use the normal approximation to find the probability that more than 48% of the people in the sample have high blood pressure?

np = 37*0.2 = 7.4 < 10

So not appropriate.

It is not appropriate to use the normal curve, since np = 7.4 < 10.

Part 2:

Now n = 82, 82*0.2 = 16.4 > 10, so ok

Mean and standard deviation:

By the Central Limit Theorem,

Mean $\mu = p = 0.2$

Standard deviation $s = \sqrt{\frac{0.2*0.8}{82}} = 0.0442$

Find the probability that more than 32% of the people in this sample have high blood pressure.

This probability is 1 subtracted by the pvalue of Z when X = 0.32. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.32 - 0.2}{0.0442]$

$Z = 2.72$

$Z = 2.72$ has a pvalue of 0.9967.

1 - 0.9967 = 0.0033

The probability that more than 32% of the people in this sample have high blood pressure is 0.0033 = 0.33%.

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3 years ago

Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.

Evgesh-ka [11]

A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420$

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

A and C aren't independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

B and C aren't independent

8 0

3 years ago