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3 years ago

A circle has a circumference of 6. It has an arc of length 1.

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

6 0

Answer:

60 degrees

Step-by-step explanation:

s = r (theta)

(theta) = s/r

= 1 / (3/π) = π/3

as a degree π/3 x 180/π = 60

60 degrees

*Circumference = 2πr

6 = 2πr

r = 6/2π = 3/π

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Part A From the choices provided, select all the questions that represent statistical questions. What was the temperature at 5:0

MArishka [77]

Answer:

Step-by-step explanation:

1. Categorical (Boston)

2. Categorical (Smoothie)

3. Statistical (States)

4. Categorical (Lunch)

5. Statistical (Library)

6 0

3 years ago

Mallorie has $5 in her wallet.if this is 10% of her monthly allowance what is her monthly allowance

marta [7]

5=.1*x
divide both side by .1
5/.1=x
x=50 dollars

4 0

3 years ago

Read 2 more answers

Zigmanuir [339]

Answer:

The weight of the water in the pool is approximately 60,000 lb·f

Step-by-step explanation:

The details of the swimming pool are;

The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet

The depth of the pool = 5 feet

The density of the water in the pool = 60 pounds per cubic foot

From the question, we have;

The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g

The volume of water in the pool = Cross-sectional area × Depth

∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³

Acceleration due to gravity, g ≈ 32.09 ft./s²

∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N

266,196.089 N ≈ 60,000 lb·f

The weight of the water in the pool ≈ 60,000 lb·f

6 0

2 years ago

Fill in the missing information. Tim Worker is doing his budget. He discovers that the average miscellaneous expense is $45.00 w

nlexa [21]

Answer:

$P(38.6$

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

We can find the probability of interest using the normal standard table and with the following difference:

$P(-0.4$

Step-by-step explanation:

Let X the random variable who represent the expense and we assume the following parameters:

$\mu = 45, \sigma 16$

And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

$P(38.6$

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

We can find the probability of interest using the normal standard table and with the following difference:

$P(-0.4$

6 0

3 years ago

Please help!!

garik1379 [7]

Given:

The radius of the cone = 14 ft

The slant height of the cone = 27 ft

To find the lateral surface area and the total surface area of the given cone.

Formula

The lateral surface area of the cone is

$LSA = \pi rl$ and

The total surface area of the cone is

$TSA = \pi r^{2} + \pi r l$

where,

r be the radius and

l be the slant height of the cone.

Now

Taking r = 14 and l = 27 we get

$LSA = \pi (14)(27)$ sq ft

or, $LSA = 1187.5$ sq ft

Again,

$TSA = \pi(14^2)+\pi (14)(27)$ sq ft

or, $TSA = 1803.3$ sq ft

Hence,

The lateral surface area of the cone is 1187.5 sq ft and the total surface area of the cone is 1803.3 sq ft. Option C.

5 0

3 years ago