Solve the equation. 72 = 3f. F=?

A room has a length of 8 m. A scale diagram is drawn of the room. In the diagram, the room has a length of 1 cm. What is the sca

liberstina [14]

The scale of the diagram as described in the task content is; 1: 8 where k= 8.

<h3>What is the scale of the diagram?</h3>

It follows from the task content that the original length of the room in discuss is 8cm. It therefore follows that the length of the room in the scale diagram is 1cm. On this note, the scale of the diagram in discuss is; 1 : 8 as 1cm on the diagram corresponds to 8cm actual length.

Read more on scale of diagrams;

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8 0

2 years ago

A where and a cylinder have the same radius and height . The volume of the cylinder is 48 cm 3. What is the volume of the sphere

SOVA2 [1]

Given:

Sphere and cylinder have same radius and height.

Volume of the cylinder = 48 cm³

To find:

The volume of the sphere.

Solution:

Radius and height of cylinder are equal.

⇒ r = h

Volume of cylinder:

$V=\pi r^2h$

Substitute the given values.

$48=\pi r^2r$ (since r = h)

$48=\pi r^3$

$48=3.14 \times r^3$

Divide by 3.14 on both sides.

$$\frac{48}{3.14} =\frac{3.14\times r^3}{3.14}$

$$15.28=r^3$

Taking cube root on both sides, we get

2.48 = r

The radius of the cylinder is 2.48 cm.

Sphere and cylinder have same radius and height.

Volume of sphere:

$$V=\frac{4}{3} \pi r^3$

$$V=\frac{4}{3} \times 3.14 \times (2.48)^3$

$V=63.85$

The volume of the sphere is 63.85 cm³.

6 0

4 years ago

Factor. 2x2 + 4x + 18

Rudiy27

You can't.

It’s already been factored down.

6 0

3 years ago

Mr. Green teaches mathematics and his class recently finished a unit on statistics. The student scores on this unit are: 40 47 5

Harrizon [31]

Answer:

Mean = 64.46, Median = 62 and Mode = Bi-modal (50 and 62)

Range of the data is 55.

Step-by-step explanation:

We are given that Mr. Green teaches mathematics and his class recently finished a unit on statistics.

<u>The student scores on this unit are:</u> 40, 47, 50, 50, 50, 54, 56, 56, 60, 60, 62, 62, 62, 63, 65, 70, 70, 72, 76, 77, 80, 85, 85, 95.

We know that Measures of Central Tendency are: Mean, Median and Mode.

Mean is calculated as;

Mean = $\frac{\sum X}{n}$

where $\sum X$ = Sum of all values in the data

n = Number of observations = 24

So, Mean = $\frac{40+ 47+ 50+ 50+ 50+ 54+ 56+ 56+ 60 +60+ 62+ 62+ 62+ 63+ 65+ 70+ 70+ 72+ 76+ 77+ 80+ 85+ 85+ 95}{24}$

= $\frac{1547}{24}$ = 64.46

So, mean of data si 64.46.

For calculating Median, we have to observe that the number of observations (n) is even or odd, i.e.;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

Now here in our data, the number of observations is even, i.e. n = 24.

So, Median = $\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(\frac{24}{2})^{th}\text{ obs.} +(\frac{24}{2}+1)^{th}\text{ obs.} }{2}$

= $\frac{(12)^{th}\text{ obs.} +(13)^{th}\text{ obs.} }{2}$

= $\frac{62 + 62 }{2}$ = $\frac{124}{2}$ = 62

Hence, the median of the data is 62.

A Mode is a value that appears maximum number of times in our data.

In our data, there are two values which appear maximum number of times, i.e. 50 and 62 as these both appear maximum 3 times in the data.

This means our data is Bi-modal with 50 and 62.

The Range is calculated as the difference between the highest and lowest value in the data.

Range = Highest value - Lowest value

= 95 - 40 = 55

Hence, range of the data is 55.

5 0

4 years ago

If JO = 32 is JKO = PMN? If so, find JN

Bas_tet [7]

The answer is B. Yes, JKO = PMN by AAS; JN = 51.

4 0

3 years ago

Read 2 more answers

Solve the equation. 72 = 3f. F=?​

Solve the equation. 72 = 3f. F=?