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3 years ago

Can anyone find the circumference in terms of pi?

Mathematics

2 answers:

gladu [14]3 years ago

7 0

Answer:

The circumference of circle is 20π cm.

Step-by-step explanation:

Given that the formula of Circumference is C = 2×π×r where r represents radius. So you have to substitute the following values into the formula :

$c = 2 \times \pi \times r$

$let \: r = 10$

$c = 2 \times \pi \times 10$

$c = 20\pi \: cm$

qaws [65]3 years ago

5 0

Answer:

20 $\pi$

Step-by-step explanation:

The formula for circumference is 2 $\pi$ r, where r stands for radius. Therefore, the circumference of this circle can be defined by:

2 $\pi$ r

2 $\pi$ 10

20 $\pi$

Hope this helps :) Please contact me if you have any questions!

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What is the next term of 54, 36, 24

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Step-by-step explanation:

54-36=18

36-24= 12

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3 years ago

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Please help this is like 40% of my Grade!!

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120

Step-by-step explanation:

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3 years ago

Frank invests $1,000 in simple interest investment account that pays 8% a year. After a number of years, he withdraws his balanc

MAXImum [283]

There are 15 years was his money invested.

<h3>What is simple interset?</h3>

Simple Interest is an easy method of calculating the interest for a loan/principal amount.

Given

Frank invests $1,000 in simple interest investment account that pays 8% a year.

After a number of years, he withdraws his balance of $2,200.

The number of years was his money inversted is;

$\rm I= P\times r \times t$

Where p is the principal amount, r is the rate of interest, I is the amount and t is the time.

Substitute all the values in the formula;

$\rm I= P\times r \times t\\\\2200-1000= 1000 \times .08 \times t\\\\1200 = 80t\\\\t = \dfrac{1200}{80}\\\\t=15$

Hence, in the 15 years was his money invested.

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2 years ago

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specifica

Natalija [7]

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal( $\mu=0.753,\sigma^{2} =0.004^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = population mean = 0.753 inch

$\sigma$ = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X $\leq$ 0.75 inch)

P(X < 0.753) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.753-0.753}{0.004} } }$ ) = P(Z < 0) = 0.50

P(X $\leq$ 0.75) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.75-0.753}{0.004} } }$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

= 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X $\leq$ 0.74 inch)

P(X < 0.75) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.75-0.753}{0.004} } }$ ) = P(Z < -0.75) = 1 - P(Z $\leq$ 0.75)

= 1 - 0.7734 = 0.2266

P(X $\leq$ 0.74) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.74-0.753}{0.004} } }$ ) = P(Z $\leq$ -3.25) = 1 - P(Z < 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

P(X > 0.76) = P( $\frac{X-\mu}{\sigma} } }$ > $\frac{0.76-0.753}{0.004} } }$ ) = P(Z > -1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

P(X < 0.74) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.74-0.753}{0.004} } }$ ) = P(Z < -3.25) = 1 - P(Z $\leq$ 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

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3 years ago

PLease help

Mariulka [41]

Answer:

the brain performs at its optimum level

Step-by-step explanation:

answer on edg 2020

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2 years ago

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