All categories

3 years ago

Find the Area of the figure below, composed of a parallelogram and two semicircles. Round to the nearest tenths place.

Mathematics

1 answer:

svet-max [94.6K]3 years ago

3 0

Answer:

100.3 units²

Step-by-step explanation:

✅Find the area of the parallelogram:

Area of Parallelogram is given as base × height

base = 10

height = 5

Area of Parallelogram = 10*5 = 50

✅Next, find the area of the two semicircles:

The two semicircles = 1 circle

Area of circle = πr²

r = 8/2 = 4

Area = π*4² = 16π = 50.3 (nearest tenth)

✅Area of figure = 50 + 50.3 = 100.3 units²

You might be interested in

Divide the decimal 134.55 by the decimal 0.5

lina2011 [118]

Answer:

269.1

Step-by-step explanation:

8 0

3 years ago

Gradient of line AB (3,7) and (5,11)

larisa [96]

Answer:

Gradient is change in y over change in x.

gradient = y1-y2 ÷ x1-x2

=7-11 ÷ 3-5

=-4 ÷ -2

4 0

3 years ago

Read 2 more answers

12/39 in simplest form is 4/13 what are two ways to get there

arlik [135]

Divide 3 on the numerator and denominator.

Hoped this helped.

~Bob Ross®

7 0

3 years ago

Read 2 more answers

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago

The maximum value of the function: f(x)= -5 x ^2 +30x-200 is?

VARVARA [1.3K]

Answer:

$\displaystyle - 155$

Step-by-step explanation:

we are given a quadratic function

$\displaystyle f(x) = - 5 {x}^{2} + 30x - 200$

we want to figure out the minimum value of the function

to do so we need to figure out the minimum value of x in the case we can consider the following formula:

$\displaystyle x _{ \rm min} = \frac{ - b}{2a}$

the given function is in the standard form i.e

$\displaystyle f(x) = a {x}^{2} + bx + c$

so we acquire:

b=30
a=-5

thus substitute:

$\displaystyle x _{ \rm min} = \frac{ - 30}{2. - 5}$

simplify multiplication:

$\displaystyle x _{ \rm min} = \frac{ - 30}{ - 10}$

simply division:

$\displaystyle x _{ \rm min} = 3$

plug in the value of minimum x to the given function:

$\displaystyle f (3)= - 5 {(3)}^{2} + 30.3 - 200$

simplify square:

$\displaystyle f (3)= - 5 {(9)}^{} + 30.3 - 200$

simplify multiplication:

$\displaystyle f (3)= - 45 + 90- 200$

simplify:

$\displaystyle f (3)= - 155$

hence,

the minimum value of the function is -155

5 0

2 years ago