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Whitepunk [10]
3 years ago
9

6. The base of a triangular field is three times its height. If the cost of cultivating the field

Mathematics
1 answer:
dedylja [7]3 years ago
7 0

Answer:

base = 519.62, height = 173.21 m

Step-by-step explanation:

Let the base and height of the triangle be represented by b and h respectively.

Thus,

b = 3h

Area of a triangle = \frac{1}{2} x base x height

For the given triangle,

area = \frac{1}{2} x b x 3h

       = \frac{3}{2}bh

Area of the triangle = \frac{3}{2}bh

To determine the number of hectares,

36 per hectare = 486

hectare = \frac{486}{36}

             = 13.5

numbers of hectares = 13.5

Area of the hectares = number of hectares x 10 000 m²

                                   = 13.5 x 10 000

                                  = 135000

Total area of the hectares = 135 000 m²

So that,

area of the hectares = area of the triangle

area of the triangle = \frac{3}{2}bh

135 000 = \frac{3}{2}bh

270000 = 3bh

bh = \frac{270000}{3}

     = 90000

bh = 90000

But, b = 3h

3h x h = 90000

3h^{2} = 90000

h^{2}  = \frac{90000}{3}

     = 30000

h = \sqrt{30000}

  = 173.2051

h = 173.21 m

So that,

b = 3 x 173.2051

  = 519.6153

b = 519.62

Therefore, the base of the triangle is 519.62 m, while the height is 173.21 m.

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Deffense [45]

Answer:

rate = 0.4 %

Step-by-step explanation:

Interest = principal x rate x time

33549 = 8635 x r x 11

33549 = 94985r

r = 33549/94985

r = 0.3532

r = 0.4

rate = 0.4 %

3 0
2 years ago
A screwdriver is 20 centimeters long, and a piece of wire is 10 millimeters long. The screwdriver is how many times as long as t
Novay_Z [31]

Answer: 20 times

Step-by-step explanation:

Given: The length of piece of wire = 10 millimeters

The length of screwdriver = 20 centimeters

We know that 1 cm = 10 mm

Therefore, the  length of screwdriver = 20\times10=200 millimeters

The ratio of the length of screwdriver and length of wire is given by :-

\frac{\text{length of screwdriver}}{\text{length of piece of wire}}=\frac{200}{10}=20

The screwdriver is 20 times as long as the wire.

8 0
3 years ago
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Calculate the circumference and area of this circle.​
Semenov [28]
Circumference
formula2TTR
0.5x2x3.14
circumference=3.14
area
formula =TT R*2
0.5*2x3.14
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3 0
2 years ago
The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q
sp2606 [1]

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.

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Mr or miss its blurd

in Tagalog: Malabo po

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2 years ago
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