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Virty [35]
3 years ago
8

Kiran scored 223 fewer points in a computer game than Tyler. Kiran scored 409 points. Which equation would Kiran use to determin

e how many points, t, Tyler scored?
Mathematics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

The equation is:

409 = T - 223

And the solution is:

T = 632

Step-by-step explanation:

Given

Represent Kiran with K and Tyler with T

Kiran score is represented as:

K = 409

Also, Kiran scored 223 less than Tyler.

This is represented as:

K = T - 223

Substitute 409 for K

409 = T - 223

Make T the subject

T = 409 +223

T = 632

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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
Rewrite in a simplest exponential form. Please show work.
Mama L [17]

(-4n^(2))*  (5n^(7)):-20n to the power of 9

(2x)^(3)*(5x^(4 ))^(2): 200x to the power of 11

4 0
3 years ago
I need help on this question can anyone please help me
MAVERICK [17]

9514 1404 393

Answer:

  p = -9 5/9

Step-by-step explanation:

Undo what is done to p.

Multiply by 2.

  p +10 = 4/9

Subtract 10.

  p = 4/9 -10

  p = -9 5/9

8 0
3 years ago
How do i do this pls help me
Delvig [45]

Answer:

y =  \frac{1}{7} x + 7

Step-by-step explanation:

x-7y = -49

i) move x to the right-hand side and change its sign

-7y = -49-x

ii) change the signs on both sides of the equation

7y = 49+x

iii) divide both sides of the equation by 7

\frac{7y}{7}  = \frac{49}{7}  +  \frac{x}{7}

y = 7 +  \frac{1}{7}  x

iv) use the commutative property to reorder the terms

* commutative property: a+b = b+a

y =  \frac{1}{7} x + 7

3 0
3 years ago
3.The side of a square frame is (2b - 1) inches. Find its area.
sweet-ann [11.9K]

Answer:

Step-by-step explanation:

(a - b)² = a² + 2ab + b²

Area of square frame =side²

                                   = (2b - 1)²

                                   = (2b)² - 2*2b * 1 + 1²

                                   = 4b² - 4b + 1

5 0
3 years ago
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