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Gemiola [76]
1 year ago
9

Work out the circumference of this circle.

Mathematics
1 answer:
DIA [1.3K]1 year ago
8 0

The circumference of a circle will be 131.964.

What is circumference of a circle?

The measurement of the circle's boundaries is called as the circumference or perimeter of the circle. whereas the circumference of a circle determines the space it occupies. The circumference of a circle is its length when it is opened up and drawn as a straight line. Units like cm or unit m are typically used to measure it. The circle's radius is considered while calculating the circumference of the circle using the formula. As a result, in order to calculate the circle's perimeter, we must know the radius or diameter value.

The radius of the circle given is = 21cm

circumference is 2πr = 2*3.142*21 = 131.964cm.

Hence the circumference of a circle 131.964cm.

Learn more about circumference of a circle, by the following link.

brainly.com/question/17380589

#SPJ1

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Marat540 [252]

Answer:

= 3√3 + 3i

Step-by-step explanation:

Cos 30° = √3/2

Sin 30°  = 1/2

Therefore;

z = 6 (cos 30° + i6 sin 30°)

  = 6 (√3/2) + i 6 (1/2)

   = 3√3 + 3i

6 0
3 years ago
Write an inequality for the following statement. A is greater than 7
yarga [219]

Answer: A>7

Step-by-step explanation:

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3 years ago
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2(x-1)-y=-8 <br> y=2x+8 <br> substitution method
BartSMP [9]

Answer:

No Solution.

Step-by-step explanation:

2(x-1)-y=-8

y=2x+8

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2x-2-(2x+8)=-8

2x-2-2x-8=-8

2x-2x-2-8=-8

0-2-8=-8

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3 0
3 years ago
a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was
Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

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

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

    P(\bar X \leq 29 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 29-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z \leq 1.60) = 0.94520

                                                                    

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0
3 years ago
A football player is running the length of a 100 yard long football field. For the first part of the field, he runs at a rate of
notka56 [123]

Answer:

\frac{2}{5}th part of the field is covered in the second sprint.

Step-by-step explanation:

A football player is running the length of a 100 yard long football field.

Let player sprints x yards with the speed = 2 yards per second.

So time taken to cover x yards player will take time = \frac{x}{2} seconds

Now rest distance (100 - x) yards when covered with the speed of 4 yards per second, so time taken to cover this distance = \frac{Distance}{Speed}

= \frac{100-x}{4} seconds

Now total time taken by the player can be represented by the equation

\frac{x}{2}+\frac{100-x}{4}=40

Now we can solve this equation for the value of x.

\frac{2x+100-x}{4}=40

x + 100 = 40×4

x + 100 = 160

x = 160 - 100 = 60 yards

And length of the second part will be = 100 - 60 = 40 yards

Now the fraction of the field covered by the player in second sprint will be

= \frac{40}{100}

= \frac{2}{5} or 40%

Therefore, \frac{2}{5}th part of the field was covered in second sprint.

5 0
3 years ago
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