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3 years ago

The angle made by the ladder with the ground is degrees, and the length of the ladder is inches.

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

7 0

Answer:

59°

58.3 inches

Step-by-step explanation:

Here is the full question :

A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground. The angle made by the ladder with the ground is degrees, and the length of the ladder is inches.

Please check the attached image for a diagram explaining this question

The angle the ladder makes with the ground is labelled x in the diagram

To find the value of x given the opposite and adjacent lengths, use tan

tan⁻¹ (opposite / adjacent)

tan⁻¹ (50 / 30)

tan⁻¹ 1.667

= 59°

the length of the ladder can be determined using Pythagoras theorem

The Pythagoras theorem : a² + b² = c²

where a = length

b = base

c = hypotenuse

√(50² + 30²)

√(2500 + 900)

√3400

= 58.3 inches

aleksley [76]3 years ago

4 0

Answer:

59.04°

58.31 inches

Step-by-step explanation:

The solution triangle is attached below :

Since we have a right angled triangle, we can apply trigonometry to obtain the angle ladder makes with the ground;

Let the angle = θ

Tanθ = opposite / Adjacent

Tanθ = 50/30

θ = tan^-1(50/30)

θ = 59.036°

θ = 59.04°

The length of ladder can be obtained using Pythagoras :

Length of ladder is the hypotenus :

Hence,

Hypotenus = √(adjacent² + opposite²)

Hypotenus = √(50² + 30²)

Hypotenus = √(2500 + 900)

Hypotenus = 58.309

Length of ladder = 58.31 inches

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A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.1 in. Determine the minimum sample

leva [86]

Answer:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (2)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025,0,1)", and we got $z_{\alpha/2}=1.96$ , replacing into formula (2) we got:

$n=(\frac{1.96(0.25)}{0.1})^2 =24.01$

So the answer for this case would be n=25 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=0.25$ represent the population standard deviation

n represent the sample size (variable of interest)

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (2)

$n=(\frac{1.96(0.25)}{0.1})^2 =24.01$

So the answer for this case would be n=25 rounded up to the nearest integer

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4 years ago

Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3227.9 g and a standa

sasho [114]

Answer:

The weight that is more extreme is the male which is -2.62

Step-by-step explanation:

Consider the provided information.

Now use the z scores formula:

$z=\frac{x-\mu}{\sigma}$

Where μ is mean and σ is the standard deviation.

For newborn males have weights with a mean of 3227.9 g and a standard deviation of 659.9 g.

Substitute the respective values in the above formula.

$z=\frac{1500-3227.9}{659.9}\approx-2.62$

For Newborn females have weights with a mean of 3073.1 g and a standard deviation of 806.8 g.

Substitute the respective values in the above formula.

$z=\frac{1500-3073.1}{806.8}\approx-1.95$

So, the more extreme of these is the male which is -2.62

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3 years ago

What is used to locate points in the coordinate plane

Maru [420]

The x axis and the y axis

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4 years ago

1. The length of a ribbon is 5/6 meter. Shayla needs pieces measuring 1/5 meter for an art project. What is the Greatest number

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Answers:

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Part 2: 4/6 m

Part 3: $5 \frac{1}{6}$

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Step By Step Explanation:

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3 years ago

The average watermelon weighs 8 lbs with a standard deviation of 1.5. Find the probability that a watermelon will weigh between

Andrew [12]

Answer:

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = 0.5932

Step-by-step explanation:

Step 1:-

by using normal distribution find the areas of given x₁ and x₂

Given The average watermelon weighs 8 lbs

μ = 8

standard deviation σ = 1.5

I) when x₁ = 6.8lbs and μ = 8 and σ = 1.5

$z_{1} = \frac{x_{1} -mean}{S.D} = \frac{6.8-8}{1.5} = - 0.8$

ii) when x₂ = 9.3 lbs and μ = 8 and σ = 1.5

$z_{2} = \frac{x_{2} -mean}{S.D} = \frac{9.3-8}{1.5} = 0.866>0$

<u>Step2</u>:-

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = A(z₂) - A(-z₁)

= A(0.866) - A(-0.8)

= A(0.866)+ A(0.8)

check below normal table

= 0.3051 + 0.2881

= 0.5932

<u>Conclusion</u>:-

The probability that a watermelon will weigh between 6.8 lbs and 9.3 lbs.

P(6.8 ≤X≤9.3) = 0.5932

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3 years ago