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3 years ago

Televisions and monitors come in two common aspect ratios, 4:3 and 16:9 (sometimes

Mathematics

1 answer:

kati45 [8]3 years ago

4 0

Answer:

The dimension of the TV = 11.2 inches × 8.4 inches

Step-by-step explanation:

Let x be the length of the TV

Let y be the width of the TV

x : y = 4:3

$\frac{x}{y} = \frac{4}{3} \\cross-multiplying\\3x = 4y \\x = \frac{4}{3} y- - - - - (1)$

From the figure attached:

using Pythagoras theorem

x² + y² = 42²

where:

$x = \frac{4}{3} y\\(\frac{4}{3}y)^2 + y^2 = 42^2\\\frac{16}{9}y^2 + y^2 = 1764\\ \frac{16y^2}{9} + \frac{y^2}{1} = 1764\\\frac{16y^2 +9y^2}{9}= 1764\\ cross-multiplying\\16y^2 + 9y^2 = 1764\\25y^2 = 1764\\y^2 = \frac{1764}{24}\\y^2 = 70.56\\y = \sqrt{70.56} \\y = 8.4\ inches\\Finding\ x:\\x = \frac{4}{3} y\\x = \frac{4}{3} \times 8.4\\ x = 1.33 \times 8.4\\x = 11.2\ inches\\\therefore\ the\ dimensions\ are\ 11.2\ inches \times 8.4\ inches$

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In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped

erica [24]

Answer:

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 24 - 1 = 23

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$ . So we have T = 2.5

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.5\frac{2}{\sqrt{24}} = 1.02$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.02 = $40.98.

The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

7 0

3 years ago

Question 4 (3 points)

trapecia [35]

The sum of the angles in a triangle is 180°, If I remember correctly.

7 0

2 years ago

On a map two cities are 6 3/4 inches apart. The scale of the map is 1/4 inch= 10 miles. What is the actual distance between the

Murljashka [212]

27 times 10 = 270 miles apart.

3 0

3 years ago

In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

4 0

3 years ago

In 2008, the total trade between the United States and Japan was $2.04 x 101^1. The total trade between the U.S. And Australia w

jolli1 [7]

Answer:

6.21 times more

Step-by-step explanation:

Assuming 2.04 x 10^11 was only a typo

you would need to first equal the powers of 10

so 20.4 x 10^10 and 3.28 x 10^10

now that they have the same powers

$\frac{20.4 * 10^{10}}{3.28 * 10^{10}} = \frac{20.4}{3.28} = 6.219512...$

if it's asking for how many times i would approximate it to 6.21 times more

3 0

3 years ago