A rectangle is removed from a right triangle to create the shaded region shown below. Find the area of the shaded region.

How do you find the formula of a half circle?

Alik [6]

Full circle
Circumference = 2*pi * r
Circumference = pi * d

Area = pi r * r = pi * r^2
Area = pi d^2 / 4

Half circle
Circumference = pi * r
Circumference = pi * r / 2

Area = pi r^2 / 2
Area = pi d^2/8

The answer depends on what you are looking for and what you are given. But those are the only common formulas I can think of at the moment.

4 0

3 years ago

Read 2 more answers

How many book marks will he make with 9 feet of ribbon

Lubov Fominskaja [6]

9 ÷ 1/3 = 27

27 bookmarks

6 0

3 years ago

Read 2 more answers

Please help this is like 40% of my Grade!!

svet-max [94.6K]

Answer:

120

Step-by-step explanation:

supplementary means the angles add up to 180

4 0

3 years ago

Totsakan school is selling tickets to a choral performance on the first day of ticket sales the school sold 10 senior tickets an

ohaa [14]

Given :

On the first day of ticket sales the school sold 10 senior tickets and 1 child ticket for a total of $85 .

The school took in $75 on the second day by selling 5 senior citizens tickets and 7 child tickets.

To Find :

The price of a senior ticket and the price of a child ticket.

Solution :

Let, price of senior ticket and child ticket is x and y respectively.

Mathematical equation of condition 1 :

10x + y = 85 ...1)

Mathematical equation of condition 2 :

5x + 7y = 75 ...2)

Solving equation 1 and 2, we get :

2(2) - (1) :

2( 5x + 7y - 75 ) - ( 10x +y - 85 ) = 0

10x + 14y - 150 - 10x - y + 85 = 0

13y = 65

y = 5

10x - 5 = 85

x = 8

Therefore, price of a senior ticket and the price of a child ticket $8 and $5.

Hence, this is the required solution.

5 0

3 years ago

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of

iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is $\\ P(z>0.48) = P(x>533.2) = 0.3156$

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a sample mean. The distribution for sample means follows a normal distribution with mean $\\ \mu$ and standard deviation $\\ \frac{\sigma}{\sqrt{n}}$ . Mathematically

$\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

For values of the sample $\\ n \ge 30$ , no matter the distribution the data come from.

And the variable z follows a standard normal distribution, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [1]

That is

$\\ z \sim N(0, 1)$

We have a variance of 3364. That is, a standard deviation of

$\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58$

The population mean is

$\\ \mu = 530$

The sample size is $\\ n = 75$

The sample mean is $\\ \overline{x} = 533.2$

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

$\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}$

$\\ z = \frac{3.2}{\frac{58}{8.66025}}$

$\\ z = \frac{3.2}{6.69726}$

$\\ z = 0.47780$

With this value of z we can consult a cumulative standard normal table (or use some statistic program) to find the cumulative probability for z (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

$\\ P(z$

However, in the question we are asked for $\\ P(z>0.48) = P(x>533.2)$ . As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

$\\ P(z>0.48) = 1 - P(z$

Thus

$\\ P(z>0.48) = 1 - 0.68439$

$\\ P(z>0.48) = 0.31561$

Rounding to four decimal places, we have

$\\ P(z>0.48) = 0.3156$

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) $\\ P(z>0.48) = P(x>533.2) = 0.3156$ .

5 0

4 years ago