a tree casts a shadow that is 20 ft long. the angle of elevation of the sun is 29°. how tall is the tree?

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Answer 1

Answer: You can use the trig ratios. Draw a diagram to see which one is appropriate.
Tan would be used.
Tan29=x/20
20Tan29=x
Put this into your calculator and see what you get

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Helpppppp i don’t know

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Answer:

Step-by-step explanation:

180 = 2x + 10 +3x

170 = 5x

34 = x

m∠B = 3(34) = 102°

How do I solve the ratio of somthin

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Add the ratio terms to get the whole. Use this as the denominator. 1 : 2 => 1 + 2 = 3.
Convert the ratio into fractions. Each ratio term becomes a numerator in a fraction. 1 : 2 => 1/3, 2/3.
Therefore, in the part-to-part ratio 1 : 2, 1 is 1/3 of the whole and 2 is 2/3 of the whole.

Diagonals AC and BD of a rhombus ABCD meet at O . If AC=8cm and BD=6cm , find sin √OCD.

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Due to the symmetry of the rhombus; AC = 2OC and BD = 2OD. Hence, we can say that OC is 4cm and OD is 3cm. Then we can say that sin(OCD) = OC/CD [CD = sqrt(OC^2 + OD^2)].

At the end, did you mean sqrt(sin(ocd)) or sin(sqrt(ocd)). Either way you can find angle OCD by doing arcsin(OC/CD).

According to the Central Limit Theorem, a) in a sufficiently large random sample, the distribution of a random variable X, with mean μ and standard deviation σ, is a normal distribution with mean μ and standard deviation σ/sqrt(n)b) The Central Limit Theorem does not apply to heavily skewed distributions.
True or False?

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The statement is True. According to the Central Limit Theorem, in a sufficiently large random sample, the distribution of a random variable X, with mean μ and standard deviation σ, is a normal distribution with mean μ and standard deviation σ/sqrt(n).

1. The Central Limit Theorem states that the distribution of the sample means of a large sample size will approach a normal distribution, regardless of the original distribution of the population from which the sample is drawn.
2. For a sufficiently large sample size, the mean of the sample means will approach the population mean (μ) and the standard deviation of the sample means will approach the population standard deviation divided by the square root of the sample size (σ/sqrt(n)).
3. Therefore, in a sufficiently large random sample, the distribution of a random variable X, with mean μ and standard deviation σ, is a normal distribution with mean μ and standard deviation σ/sqrt(n).

This is the case because the Central Limit Theorem states that the distribution of sample means is approximately normal, regardless of the original distribution of the population from which the sample is drawn.

To know more about central limit theorem, refer here:

brainly.com/question/18403552#

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True so there is the answer

If Joel’s family starts with a full tank of gas, can they drive the car for 15 days without the warning light coming on?