Workout the value of x.

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

Can someone help me with this it is confusing :(

Marina CMI [18]

What are you wanting help with

7 0

3 years ago

Can you help me plss!

Ksivusya [100]

(1/3) × the cone's volume = The cylinder's volume.

Step-by-step explanation:

Step 1:

The volume of any cone is obtained by multiplying $\frac{1}{3}$ with π, the square of the radius ( $r^{2}$ ) and the height ( $h$ ).

So the volume of the cone, $V=\pi r^{2} \frac{h}{3}$ .

Step 2:

The cylinder's volume is nearly the same as the cone but instead by multiplying $\frac{1}{3}$ we multiply with 1.

So the cylinder's volume is determined by multiplying π with the square of the radius of the cylinder ( $r^{2}$ ) and the height of the cylinder ( $h$ ).

So the the cone's volume, $V = \pi r^{2} h$ .

Step 3:

Now we equate both the volumes to each other.

The cone's volume : The cylinder's volume = $\pi r^{2} \frac{h}{3}: \pi r^{2} h$ = $\frac{1}{3} : 1$ .

So if we multiply the cone's volume with $\frac{1}{3}$ we will get the cylinder's volume with the same dimensions.

5 0

3 years ago

5х2 +х — 4<br> Factor completely

Step2247 [10]

Answer:

(5x-4)(x+1)

Step-by-step explanation:

3 0

3 years ago

Will give brainliest give the answer and steps

zvonat [6]

Answer:

m = (p+7n)/3

Step-by-step explanation:

3m-7n = p

3m = p+7n

m = (p+7n)/3

7 0

2 years ago