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3 years ago

Plssssss help me! What is the area of the shaded portion to the nearest hundredth?

Mathematics

1 answer:

Katarina [22]3 years ago

7 0

Step-by-step explanation:

A1= area of the rectangle

A1= 4yd • 10yd

A1= 40 square yd

A2= area of the part of the circular section

A2= 180°• pi • 4 square yd/360°

A2= 180° • 16yd • pi/ 360°

A2= 16yd•pi/2

A2= 8yd•pi

A2= 25,12yd

A= areo of rhe shaded portion

A= A1-A2

A= 40yd-25,12yd

A= 14,88yd

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Is 90 1/10th of 900?

mestny [16]

Yes, 90 is 1/10 of 900

Because

90 = 1/10 x 900
90 = 90

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3 years ago

A researcher is interested in estimating the mean weight of a semi trailer truck to determine the potential load capacity. She t

Kaylis [27]

Answer: 95% confidence interval = 20,000 ± 2.12 $\times$ $\frac{1500}{\sqrt{17} }$

( 19228.736 , 20771.263 ) OR ( 19229 , 20771 )

Step-by-step explanation:

Given :

Sample size(n) = 17

Sample mean = 20000

Sample standard deviation = 1,500

5% confidence

∴ $\frac{\alpha}{2}$ = 0.025

Degree of freedom ( $d_{f}$ ) = n-1 = 16

∵ Critical value at ( 0.025 , 16 ) = 2.12

∴ 95% confidence interval = mean ± $Z_{c}$ $\times$ $\frac{\sigma}{\sqrt{n} }$

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3 years ago

Please answer and show work!!!!!!!!!

bixtya [17]

⚘Kindly refer to the attachments for complete solutions!!~

4 0

2 years ago

6 = -4x + y -5x - y = 21 Please Help Fast!!!!!! Substitution

Vera_Pavlovna [14]

Answer:

y = 4x + 6

-5x - 4x - 6 = 21

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5 0

3 years ago

Serena has attached a 10 inch ribbon to the corners of a frame to hang it on the wall. The frame 9 inches wide. How far above th

kherson [118]

Answer: The hook would be 2.2 inches (approximately) above the top of the frame

Step-by-step explanation: Please refer to the picture attached for further details.

The top of the picture frame has been given as 9 inches and a 10 inch ribbon has been attached in order to hang it on a wall. The ribbon at the point of being hung up would be divided into 5 inches on either side (as shown in the picture). The line from the tip/hook down to the frame would divide the length of the frame into two equal lengths, that is 4.5 inches on either side of the hook. This would effectively give us two similar right angled triangles with sides 5 inches, 4.5 inches and a third side yet unknown. That third side is the distance from the hook to the top of the frame. The distance is calculated by using the Pythagoras theorem which states as follows;

AC^2 = AB^2 + BC^2

Where AC is the hypotenuse (longest side) and AB and BC are the other two sides

5^2 = 4.5^2 + BC^2

25 = 20.25 + BC^2

Subtract 20.25 from both sides of the equation

4.75 = BC^2

Add the square root sign to both sides of the equation

2.1794 = BC

Rounded up to the nearest tenth, the distance from the hook to the top of the frame will be 2.2 inches

4 0

3 years ago