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2 years ago

How to find the area of this figure

Mathematics

1 answer:

stiv31 [10]2 years ago

3 0

Answer:

45 square units.

Step-by-step explanation:

Applying the pythagoras theorem:

Base of the bigger triangle = $\sqrt{15^2 - 9^2}$ = $\sqrt{144}$ = 12 units

Area of a triangle is given by $\frac{1}{2}$ × base × height

The area of the bigger triangle is $\frac{1}{2}$ × 12 × 9 = 54 square units.

The area of the smallest triangle is $\frac{1}{2}$ × (12 - 10) × 9 = 9 square units.

The area of the figure( required triangle) is 54 - 9 = 45 square units.

You might be interested in

A large emerald with a mass of 378.24 grams was recently discovered in a mine. If the density of the emerald is 2.76 grams over

stepan [7]

Answer:

137.04 cm³

Step-by-step explanation:

Mass = Density*Volume
V = m/d
V = 378.24/2.76 = 137.04 cm³ (rounded)

5 0

2 years ago

Read 2 more answers

The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g

Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about $\\ P(z>1.56) = 0.0594$ .

Step-by-step explanation:

This is a problem of the distribution of sample means. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves normally for samples sizes equal or greater than 30 $\\ n \geq 30$ . Mathematically

$\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}})$ [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a normal standard distribution, i.e., a normal distribution that has a population mean $\\ \mu = 0$ and a population standard deviation $\\ \sigma = 1$ .

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

From the question, we know that

The population mean is $\\ \mu = 88$ WPM
The population standard deviation is $\\ \sigma = 14$ WPM

We also know the size of the sample for this case: $\\ n = 137$ sixth graders.

We need to estimate the probability that a sample mean being greater than $\\ \overline{X} = 89.87$ WPM in the distribution of sample means. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}$

$\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}$

$\\ Z = 1.5634 \approx 1.56$

This is a standardized value and it tells us that the sample with mean 89.87 is 1.56 standard deviations above the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any cumulative standard normal table available in Statistics books or on the Internet. Of course, this probability is the same that $\\ P(\overline{X} < 89.87)$ . Then