Ask question

Ask question

All categories

3 years ago

8

Find the area Plz help

2 answers:

svetoff [14.1K]3 years ago

7 0

The area of the given triangle is equal to half the area of a rectangle with the same dimensions. Thus, the triangle’s area is equal to 7*8/2 = 56/2 = 28.

Katena32 [7]3 years ago

5 0

To calculate the area of a triangle, multiply the base and the height, then divide the product by 2.

7*8=56

56/2=28

Area=28 feet squared

You might be interested in

The vertices of ΔXYZ are X(−3, −6), Y(21, −6), and Z(21, 4). What is the perimeter of the triangle?

insens350 [35]

$XY= \sqrt{(21-(-3))^2+(-6-(-6))^2} = \sqrt{(21+3)^2+(12-12)^2} = \\ = \sqrt{24^2+0} = \sqrt{24^2} =24 \\ YZ= \sqrt{(21-21)^2+(4-(-6))^2} = \sqrt{0^2+(4+6)^2} = \sqrt{10^2} =10 \\ XZ= \sqrt{(21-(-3))^2+(4-(-6))^2} = \sqrt{(21+3)^2+(4+6)^2} = \\ =\sqrt{24^2+10^2} = \sqrt{576+100} = \sqrt{676} =26 \\ \\ P=XY+YZ+XZ=24+26+10=60$

8 0

3 years ago

Is 0.18 a terminating, repeating, or non repeating decimal

Paha777 [63]

Answer:

a non repeating decimal

Step-by-step explanation:

answer from alexa device

6 0

4 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

4 years ago

HELP MEH PLS AND THX !!!!!!!!!!!!!!!!!!!

natta225 [31]

B!!!!!!!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago

What is the answer I need help ?

lara [203]

Answer:

4

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers