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2 years ago

What could be A equals 1/4 of 24

Mathematics

2 answers:

skad [1K]2 years ago

8 0

1/4 of 24

1 times 6= 6

4 times 6=24

so its 6

Leya [2.2K]2 years ago

3 0

If A= one quarter of 24
then 24/4=6
therefore,
A=6

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Notebooks are priced 3.20 for 4 notebooks Pencils are priced 4.80 for 12 pencils calculate the unit rate for cost per pencil. Ma

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Answer:

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Step-by-step explanation:

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2 years ago

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lara31 [8.8K]

Answer:

The area of the shaded part = 61.46

Step-by-step explanation:

Assume the hypotenuse of the triangle is c (c>0)

As the triangle inscribed in the semi circle is the right angle triangle, its hypotenuse is equal to the diameter of the circle.

The hypotenuse of the triangle can be calculated by Pythagoras theorem as following: $c^{2} =a^{2} +b^{2} =(2\sqrt{5}) ^{2} + (4\sqrt{5}) ^{2} = 20 + 80 = 100$

=> c = 10

So that the semi circle has the diameter = 10 => its radius = 5

The total area of 2 semi circles is equal to the area of the circle with radius =5

=> The total area of 2 semi circles is: $\pi$ x $5^{2}$ = 25 $\pi$

The area of a triangle inscribed in the semi circle is: 1/2 x a x b = 1/2 x $2\sqrt{5\\$ x $4\sqrt{5}$ = 20

=> The area of 2 triangles inscribed in 2 semi circles is: 2 x 20 = 40

The area of the square is: $c^{2}$ = $10^{2} =100$

It can be seen that:

The area of the shaded part = The area of the square - The total area of 2 semi circles + The total are of 2 triangles inscribed in semi circles

= 100 - 25 $\pi$ + 40 = 61.46

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2 years ago

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2 years ago

Read 2 more answers

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. The

Tems11 [23]

Answer:

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , 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 $1 - \frac{\alpha}{2}$ .

They randomly survey 387 drivers and find that 298 claim to always buckle up.

This means that $n = 387, \pi = \frac{298}{387} = 0.77$

84% confidence level

So $\alpha = 0.16$ , z is the value of Z that has a p-value of $1 - \frac{0.16}{2} = 0.92$ , so $Z = 1.405$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.74$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.405\sqrt{\frac{0.77*0.23}{387}} = 0.8$

The 84% confidence interval for the population proportion that claim to always buckle up is (0.74, 0.80).

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2 years ago