Plz help me: ) : ) plz

A total of 684 tickets were sold for the school play. They were either adult tickets or student tickets. There were 66 fewer stu

dlinn [17]

375

Mark brainliest please

6 0

2 years ago

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A tree 12m high is broken by the wind in such a way that its top touches the ground and makes an angle 60° with the ground. at w

Katena32 [7]

The height of broken part of tree from ground is 5.569m.

Justification:

Let BD is a tree of height 12 m.

Suppose it got bent at a point C and let the part CD take the position CA, meeting the ground at A.

i.e., CD = AC = h m

Broken part makes 60° angle from ground

So, ∠BAC = 60°

Now, height of remaining part of tree = (12 – h)m.

In right angled ∆ABC,

sin 60° = BC/AC

⇒ √3/2 = (12 - h)/h

⇒ √3h = 2(12 – h)

⇒ √3h = 24 – 2h

⇒ √3h + 2h = 24

⇒ h(√3 + 2) = 24

⇒ h(1.732 + 2) = 24

⇒ h(3.732) = 24

⇒ h = 24/3.732 = 6.4308 m

Hence, height of broken tree from ground

⇒ BC = 12 – h

⇒ 12 – 6.4308 = 5.569m

Hence, tree is broken 5.569 m from ground.

Note: See attached picture.

4 0

3 years ago

If D = w2 - 7 and C = 3 + 10w, find an expression that equals D C in standard form.

solniwko [45]

W^2-10w-10

Step-by-step explanation:

D=w^2-7

C=3+10w

D-C

(W^2-7)-(3+10w)

W^2-7-3-10w

7 0

2 years ago

Find the unit rate kenny reads 5/8 pages in 2/3 minutes

Cerrena [4.2K]

.9375 pages / 1 minute

or 15/16 pages for every 16/16 minute.

4 0

4 years ago

Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbic

jolli1 [7]

Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

$p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247$