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3 years ago

6

Mathematics

2 answers:

zloy xaker [14]3 years ago

7 0

Solution:

Note that:

Area of circle: πr²
Circle/2 = Semicircle
Area of semicircle = πr²/2
Radius = 6 units

Finding the area of the semicircle:

πr²/2
=> 3.14 x 6 x 6/2
=> 3.14 x 6 x 3
=> 56.52 units²

We can conclude that the area of the semicircle is 56.52 units².

SVETLANKA909090 [29]3 years ago

6 0

Answer:

56.52 units²

explanation:

We should know circle's area: πr²
Then the semi circle shall have half area, so formula: (1/2)πr²

Steps:

Using the formula:

→ (1/2)πr²

→ (1/2)(3.14)(6)²

→ 56.52 units²

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The first person to answer this gets brainliest

shusha [124]

Answer:

Step-by-step explanation:

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2 years ago

Thirty percent of the students in a high school face a disciplinary action of some kind before they graduate. Of those students,

BaLLatris [955]

Answer:

54% of the students from the high school go on to college

Step-by-step explanation:

We have these following percentages:

30% of the students from a high school face a disciplinary action of some kind before they graduate. Of those, 40% go on to college.

70% do not face action, and of those, 60% go to college.

What percent of the students from the high school go on to college?

Sum of 40% of 30%(face disciplinary action) and 70% of 60%(do not

face disciplinary action). So

$0.4*0.3 + 0.7*0.6 = 0.12 + 0.42 = 0.54$

54% of the students from the high school go on to college

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3 years ago

Simplify each expression: a. 32=(-7+5) b. 18+42 =(-8)

Sedbober [7]

Unfortunately these aren’t expressions, these are equations that are not equivalent.

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3 years ago

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T

Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. In this problem, $\sigma = 0.25$

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

$n = 96.04$

Rounding up, 97

8 0

3 years ago

12x -20=n (3x-5) I need help I don’t get it please help me.

Salsk061 [2.6K]

Answer:

The value of n = 4 and

The value of x = $\frac{5}{3}$

Step-by-step explanation:

Method I:

As given, 12x -20=n (3x-5) .........(1)

Firstly , take 12x - 20

As 12x - 20 can be written as 4(3x - 5)

∴ equation (1) becomes

4(3x - 5) = n (3x-5)

Divide the equation by (3x-5) , we get

4 = n

Method II:

As given, 12x -20=n (3x-5)

⇒12x - 20 = 3nx - 5n

⇒12x - 3nx = 20 - 5n

⇒3x(4 - n) = 5( 4 - n)

Divide the equation by (4-n) , we get

3x = 5

⇒ x = $\frac{5}{3}$

∴ we get

n = 4 and x = $\frac{5}{3}$

Now, check that value of n and x satisfies the equation or not

12( $\frac{5}{3}$ ) - 20 = 4( 3( $\frac{5}{3}$ ) - 5)

⇒4(5) - 20 = 4(5 - 5)

⇒20 - 20 = 4(0)

⇒ 0 = 0

Hence , we get

The values of x and n satisfies the equation

so, we get

n = 4 and x = $\frac{5}{3}$

5 0

3 years ago