All categories

3 years ago

PLEASE, NO LINKS OR FILES. If the diameter of a cylinder is 6 yards, then the value of r 2 is 9 square yards. True False

Mathematics

2 answers:

klio [65]3 years ago

6 0

Answer:

True.

Step-by-step explanation:

R² is radius to the second power. You were given the diameter, so the radius will be half of that diameter.

3 to the second power, (or 3 x3), is 9.

Nina [5.8K]3 years ago

5 0

Answer:

True

Step-by-step explanation:

Given data

Diameter =6 yards

Radius R= Diameter/2= 6/2 = 3 yards

R²=3*3=9square yards

Hence R^2 = 9square yards= True

You might be interested in

Dale finished his English assignment in 1/3 hours . Then he completed his chemistry assignment in 1/2 hours . How much more time

11Alexandr11 [23.1K]

Answer:

1/6 hours

Step-by-step explanation:

Since it wants a fraction form all you have to do is subract 1/3 from 1/2

1/2 - 1/3

cross multiply:

3/2-2/3

subtract the numerator but multiply the denominator and simplify if you have to:

1/6

6 0

3 years ago

A patient receives 150 cc of medication in an IV drip over 4 hours. What is the rate per hour of this dose

geniusboy [140]

150 cc per 4 hours divide by 4 equals 37.5 cc per hour

3 0

3 years ago

Solve for x. <br> -0.7(0.3x - 0.8) = -4.48

marta [7]

Divide both sides by -0.7 to get rid of that pesky thing in the front
0.3x-0.8=6.4
add 0.8 to both sides
0.3x=7.2
divide both sides by 0.3
x=24

8 0

3 years ago

Please help me, ASAP! Algebraic equation needed.

Simora [160]

6 × 10 = 60$ per 6 pounds
14 × 8 = 112$ per 14 pounds

you'll get 20 pounds of the candy mixture and will cost 172

172/20
8.6$ per pound

7 0

3 years ago

A large cooperation has quality control over its fertilizers. The fertilizes are composed of nitrogen. The fertilizer requires 3

JulsSmile [24]

Answer:

It is impossible that the mean of the sample of 100 fertilizers less than 2 mg.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.5 mg

Standard Deviation, σ = 0.1 mg

We are given that the distribution of percentage of nitrogen score is unknown.

Formula with sampling:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P( mean of the sample of 100 fertilizers less than 2 mg)

P(x < 2)

$P( x < 2) = P( z < \displaystyle\frac{2-2.5}{\frac{0.1}{\sqrt{100}}}) = P(z < -50)$

Calculation the value from standard normal z table, we have,

$P(x < 2) =0 = 0\%$

Thus, it is impossible that the mean of the sample of 100 fertilizers less than 2 mg.

4 0

3 years ago