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3 years ago

What is the next term of the arithmetic sequence? -2,3, 8, 13, 18,

Mathematics

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

Step-by-step explanation:

difference between number is 5

5n-7 to get to next term

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3. Jim bought 2 books yesterday. The price

natima [27]

Answer:

26.40 divided by 2 is 13.20 13.20 is your answer!

7 0

3 years ago

You roll a fair 6-sided die. what is the probability rolling greater than 4

Readme [11.4K]

Answer:

1/3

Step-by-step explanation:

There are 6 possible outcomes when you roll this die: 1,2,3,4,5 and 6. Of these, only 5 and 6 are greater than 4, which is 2 successful outcomes. Probability is successful outcomes/total outcomes = 2/6 = 1/3. Hope this helps!

6 0

3 years ago

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

Yo can someone help please

Mama L [17]

Answer: 5/3

One of the side lengths goes from 3 to 5. So, the scale factor is 5/3.

Hope that helped!

7 0

2 years ago

If it takes one minute to run 130 meters how long will it take to run 1494.23 meters

Rus_ich [418]

We have to see how many times 130 fits into 1494.23:

$1494.23/130=11.4940....$

So now we have to multiply this by the amount of time needed for 130 meters:
11.4940*1=11.4940.

So the answer is 11.4940 minutes.

4 0

3 years ago

Read 2 more answers