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4 years ago

To inscribe a hexagon inside a circle, to which length should you set your compass in order to draw the six vertices?

Mathematics

1 answer:

vesna_86 [32]4 years ago

6 0

Answer:

You need to set your compass to the radius of the circle.

Step-by-step explanation:

So therefore the answer is <u><em>A)</em></u>

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193 feet to the nearest integer

Ivenika [448]

Answer:

200 feet

Step-by-step explanation:

How to find nearest integer

Exp

59 to nearest integer

Hope this will help you :))

6 0

3 years ago

Can I get some help on this? Please?

egoroff_w [7]

Where is the question

7 0

4 years ago

Read 2 more answers

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

3 years ago

In verbal expression

AlladinOne [14]

Variable squared minus negative 2

8 0

3 years ago

Find the inverse of the following function.

Naddik [55]

For this case we must find the inverse of the following function:

$f (x) = 8 \sqrt {x}$

For this, we follow the steps below:

We change f (x) to y:

$y = 8 \sqrt {x}$

We exchange the variables:

$x = 8 \sqrt {y}$

We solve for y:

$8 \sqrt {y} = x$

We divide between 8 on both sides of the equation:

$\sqrt {y} = \frac {x} {8}$

We raise both sides of the equation to the square to remove the root:

$y = (\frac {x} {8}) ^ 2\\y = \frac {x ^ 2} {64}$

So, the inverse is: $f ^ {- 1} (x) = \frac {x ^ 2} {64}$

ANswer:

Option C

3 0

3 years ago