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Masja [62]
3 years ago
14

a wireless phone plan cost eric $35 for a month of service during which he spent 450 text messages. if he was charged a fixed fe

e of $12.50, how much did he pay per text?
Mathematics
1 answer:
Sonbull [250]3 years ago
6 0
35= 450X + 12.50 it should be .05 

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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
An investment account was opened with an initial deposit of 9,600 and earns 7.4% interest compounded continuously, how much will
emmainna [20.7K]

Answer:

10,656

Step-by-step explanation:

I=PRT

I=(9600)(0.074)(15)

Multiply those

I=10,656

3 0
2 years ago
How can I Factor 7 + 14x ??
wariber [46]

Answer: 7(1+2x)

Step-by-step explanation:

7 and 14 can be divided by 7. As a result, you take 7 out and bam!

6 0
3 years ago
Two cars leave town at the same time going in the same direction on the same road.One travels 32 mi/h and the other travels 47 m
gladu [14]
In 4.6 hours they will be 69 miles apart 
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3 years ago
Complete the table of ordered pairs for the given linear equation
nika2105 [10]

Answer:

First blank= -1

Second blank= 14

Step-by-step explanation:

Plug in x to the equation. 1/7 x (-7)= -7/7=-1

2=1/7(x) to divide fractions you multiply by the reciprocal

2 x 7/1= 1/7× 7/1

14/1=14

4 0
3 years ago
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