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Alekssandra [29.7K]
3 years ago
6

Now move the center of rotation, R, to a different position on the coordinate plane.

Mathematics
1 answer:
Scilla [17]3 years ago
4 0

Answer: If the center of rotation is changed, the pentagon maps back onto itself only one time, when the value of a is 360 degrees that is, when the pentagon completes a full rotation.  (just a heads up if your going to copy and paste, change it up a bit because this is word for word :) hope it helped!)

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Area of square = 144x <br> Side of square = 4x <br> What does X___ = please answer this !!!
trapecia [35]

Answer:

1. Area = Base * Height

2. 144x = 4x * 4x

3. 144x = 16(x^2)

4.  9x = x^2

5.  x = 9

Step-by-step explanation:

6 0
2 years ago
My last one got deleted but whats 10x5 i dont actually but u get it
andre [41]

Answer:

50 and 160

Step-by-step explanation:

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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
Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!
likoan [24]

Answer:

1017.9ft ^{3}

Step-by-step explanation:

volume \: of \: cylinder \\ v = \pi \times r {}^{2}  \times h \\  = \pi  \times 9^2 \times 4 \\  = 1017.9

4 0
2 years ago
I need help to solve this problem p/0,05p
AfilCa [17]

p / 0.05p

= 1 / 0.05

= 20  answer

7 0
3 years ago
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