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3 years ago

50 points!!!

Mathematics

1 answer:

Fynjy0 [20]3 years ago

8 0

Answer:

first I don't know what else please help me too lol.

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The parallelogram has a base of 9 cm and a high of 21 cm what is the area of the parallelogram

Assoli18 [71]

$a = b \times h$

$a = 9cm \times 21cm$

$a = 189cm ^{2}$

7 0

3 years ago

The senior class want to play a practical joke on the principal, who was a former college basketball star. They want to fill his

steposvetlana [31]

The number of basketball that will fill up the entire office is approximately 16,615.

Recall:

Volume of a spherical shape = $\frac{4}{3} \pi r^3$

Volume of a rectangular prism = $l \times w \times h$

Given:

Diameter of basketball = 9.5 in.

Radius of the ball = 1/2 of 9.5 = 4.75 in.

Radius of the ball in ft = 0.4 ft (12 inches = 1 ft)

Dimension of the office (rectangular prism) = 20 ft by 18 ft by 12 ft

First, find the volume of the basketball:

Volume of ball = $\frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times 4.75^3\\$

Volume of basketball = $448.92 $ in^3$

Convert to $ft^3$

$1728 $ in.^3 = 1 $ ft^3$

Therefore,

Volume of basketball = $\frac{448.92}{1728} = 0.26 $ ft^3$

Find the volume of the office (rectangular prism):

Volume of the office = $20 \times 18 \times 12 = 4,320 $ ft^3$

Number of basket ball that will fill the office = Volume of office / volume of basketball

Thus:

Number of basket ball that will fill the office = $\frac{4,320}{0.26} = 16,615$

Therefore, it will take approximately 16,615 balls to fill up the entire office.

Learn more here:

brainly.com/question/16098833

6 0

3 years ago

Ue the following information to complete the partial worksheet for bills company. Record the appropriate adjusting entries using

lbvjy [14]

If I am reading this right, it looks like the 10, 3, 2, 1 are Adjustments and the Adjusted TB should equal the difference. Make sure you know how to add and subtract the debit and credit adjustments correctly.

TB +/- Adj = ATB

7 0

3 years ago

What is wrong with the equation? 2 x−3 dx = x−2 −2 2 −3 = − 5 72 −3 f(x) = x−3 is continuous on the interval [−3, 2] so FTC2 can

Wewaii [24]

Answer:

Hello your question is incomplete below is the complete question

What is wrong with the equation? integral^2 _3 x^-3 dx = x^-2/-2]^2 _3 = -5/72 f(x) = x^-3 is continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous on the interval [-3, 2] so FTC2 cannot be applied. f(x) = x^-3 is not continuous at x = -3, so FTC2 cannot be applied The lower limit is less than 0, so FTC2 cannot be applied. There is nothing wrong with the equation. If f(2) = 14, f' is continuous, and f'(x) dx = 15, what is the value of f(7)? F(7) =

answer : The value of f(7) = 29

Step-by-step explanation:

Attached below is the detailed solution

Hence : F(7) - 14 = 15

F(7) = 15 + 14 = 29

8 0

4 years ago

Lower bound is 18, upper bound is 28. what is the point estimate of the population mean

hichkok12 [17]

23 is the mean estimate

7 0

4 years ago

Read 2 more answers