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3 years ago

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A right cylinder has a diameter of 40 mm and a height of 30 mm, as shown. Semester Test, Part 1 What is the volume of the cylind

er?
a. 37,680 mm3
b. 1256 mm3
c. 150,720 mm3
d. 12,000 mm3

2 answers:

sukhopar [10]3 years ago

8 0

Volume = Area * height
= (pi/4)*D^2 * h
= (pi/4)*(40)^2 * 30
= 37,680 mm3
so, the answer is (a)

lions [1.4K]3 years ago

3 0

Answer: a. $37,680\ mm^3$

Step-by-step explanation:

We know that the volume of a cylinder is given by :-

$V=\pi r^2 h$ , where r is radius and h is height of the cylinder.

Given: The diameter of cylinder = 40 mm

The radius of cylinder = $r=\dfrac{40}{2}=20\ mm$

The height of cylinder = 30 mm

Then , the volume of the cylinder will be :-

$V=(3.14) (20)^2 (30)=37680\ mm^3$

Hence, the volume of the cylinder = $V=37680\ mm^3$

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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

The formula is m=4a+2c, where c= number of children; m= number of meatballs; a = number of adults. How many meatballs are requir

lorasvet [3.4K]

M=(4x25)+(2x6)
=100+12
m=112

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4 years ago

Can anyone pls help me in math<3

FinnZ [79.3K]

Answer:

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Step-by-step explanation:

Let $\displaystyle \large{d}$ be common difference & $\displaystyle \large{x}$ be the third number. Set up the following equations:

$\displaystyle \large{x-2d=-2 \to (1)}\\\\\displaystyle \large{x+2d=26 \to (2)}$

Solve the simultaneous equation for x-term:

$\displaystyle \large{2x=24}\\\\\displaystyle \large{x=12}$

Let’s make sure that we get accurately answer. Substitute x = 12 in any equations which I’ll choose (1):

$\displaystyle \large{12-2d=-2}\\\\\displaystyle \large{-2d=-14}\\\\\displaystyle \large{d=7}$

Now add each terms with common difference as 7:

-2+7 = 5

5+7 = 12

12+7 = 19

19+7 = 26

So our pattern is -2, <u>5</u>, <u>12</u>, <u>19</u>, 26. Since the question only asks for third number then the answer is 12

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2 years ago

Write an equation in point slope form for the line with a slope of 3 passes through (1,2)

alexgriva [62]

Answer:

$y-2=3(x-1)$

Step-by-step explanation:

The point-slope form of a line is given as:

$y-y_1=m(x-x_1)$

Where

m is the slope

$x_1$ is the x-coordinate of the point given (passing through the line)

$y_1$ is the y-coordinate of the point given (passing through the line)

Now,

Given,

Slope = m = 3

$x_1$ = 1

$y_1$ = 2

Now, we simply plug these into the formula for point-slope form of a line:

$y-y_1=m(x-x_1)\\y-2=3(x-1)$

This is the point-slope form.

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3 years ago

A garden is in the shape of a rectangle 85m by 50m

Xelga [282]

Answer:

2890 sq. meters

Step-by-step explanation:

The area of the whole garden is 85 x 50 = 4250 sq. meters. Since the flowers take up 32%, that means the grass must take up 100-32, or 68% of the garden. 68% of 4250 is 2890 sq. meters.

Hope this helped! :)

7 0

4 years ago

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