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2 years ago

Find the length of chord AD

Mathematics

1 answer:

Ket [755]2 years ago

7 0

9514 1404 393

Answer:

AD = 15

Step-by-step explanation:

The products of the segment lengths are the same.

(x -2)(x +7) = 4(2x -1)

x^2 +5x -14 = 8x -4

x^2 -3x -10 = 0

(x -5)(x +2) = 0

The positive solution is x=5.

Chord AD is ...

(x -2) +(x +7) = 2x +5 = 2(5) +5 = 15

Chord AD is 15 units long.

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The answer is 9.5. Hope that helps!

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Ellie, the elephant, eats 4 times as much as Popcorn, the pony. Zac, the zebra, eats twice as much as Popcorn. Altogether, they

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Ellie requires 280 kg, Popcorn requires 70 kg and Zac requires 140 kg each week

Step-by-step explanation:

The given is:

Ellie, the elephant, eats 4 times as much as Popcorn, the pony
Zac, the zebra, eats twice as much as Popcorn
Altogether, they eat 490 kilograms of feed per week

We need to find how much feed each of them requires each week

Assume that Popcorn eats x kg each week

∵ Popcorn eats x kg each week

∵ Ellie eats 4 times as much as Popcorn

- Multiply x by 4 to find how much Ellie eats

∴ Ellie eats = 4 × x = 4 x kg

∵ Zac eats twice as much as Popcorn

- Multiply x by 2 o find how much Zac eats

∴ Zac eats = x × 2 = 2 x kg

- Add all of them to find how much they eat altogether

∴ Altogether eats = x + 4 x + 2 x = 7 x kg

∵ Altogether, they eat 490 kilograms of feed

- Equate the sum of their eat by 490

∴ 7 x = 490

- Divide both sides by 7

∴ x = 70

∵ Ellie eats 4 x

∴ Ellie eats = 4(70) = 280 kg

∵ Popcorn eats x

∴ Popcorn eats = 70 kg

∵ Zac eats 2 x

∴ Zac eats = 2(70) = 140 kg

Ellie requires 280 kg, Popcorn requires 70 kg and Zac requires 140 kg each week

Learn more:

You can learn more about the the word problems in brainly.com/question/3950386

#LearnwithBrainly

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3 years ago

MP Persevere with Problems Suppose 20

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Answer:

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If 20 of 140 people play tennis, then in a group of 504 people, we expect ...

(20/140) × 504 = 72 . . . . tennis players

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We expect 8 of the 504 people have a tennis coach. We made this prediction by applying the given rates to the larger group.

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Helen is 8 years older than Jane. Twenty years ago Helen was three times as old as Jane. How old is each now?

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Helen is 16 and Jane is 8

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A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is <u>n for which M = 0.04.</u>

There is no estimate, hence $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

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2 years ago