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UNO [17]
2 years ago
8

Work out the length of x

Mathematics
2 answers:
Ierofanga [76]2 years ago
6 0

Answer:

11.2 mm

Explanation:

Given a right angle triangle, In order to find the third side: Use Pythagoras Theorem: a² + b² = c²

Insert Values:

x² + 3.8² = 11.8²

x² + 14.44 = 139.24

x² = 139.24 - 14.44   (subtract)

x² = 124.8

x = √124.8              (square root)

x = 11.1714 ≈ 11.2     (rounded to 3 significant figures)

MakcuM [25]2 years ago
5 0
<h3>given:</h3>

3.8 mm

11.8 mm

<h3>to find:</h3>

the unknown "x"

<h3>solution:</h3>

we'll have to use pythagoras theorem.

unknown value is x

{c}^{2}  =  {b}^{2}  -  {a}^{2}

{x}^{2}  =  {11.8}^{2}  -  {3.8}^{2}

{x}^{2}  = 124.8

{x}^{2}  =  \sqrt{124.8}

x = 11.17139204

x = 11.2 \: mm

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Answer: g(1) is 1.

Step-by-step explanation:

If we are solving for g of 1  then the first equation can't be used to solve for g of 1 because it only works in x is equal to -2.

The same way the second function will not work because x has to equal to -1.

But in the last function which is written as a polynomial it will work for this situation because x is not equal to -2 or -1 so apart from those numbers every number will work the same way 1 will work.    

So plot 1 into the function and solve for it .

g(1) = 1^3 - 1^2 + 1  

g(1)=  1 - 1 + 1

g(1)= 1

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4 years ago
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A candy bar cost the store $0.79. The store wants a 65% mark up. What would the price of the candy bar be to have a 65% mark up?
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Answer:

1.30$

Step-by-step explanation:

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3 years ago
What is the smallest number with the factors 1,2,3,4, and 5? What is the special name for it?
klasskru [66]
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1.4

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
4 years ago
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