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2 years ago

6 1/5 × 5 = please help me i dont know the answer

Mathematics

2 answers:

Elan Coil [88]2 years ago

7 0

Answer:

Step-by-step explanation:

Zanzabum2 years ago

5 0

For 6 1/5
This would equal 31/5 because 5 x 6 is 30 and then 30 + 1 is 31. We keep 5 in the denominator.

So we’re left with 31/5 x 5

5 is the same as 5/1

So we technically have 31/5 x 5/1

When we multiply the two fractions, we’re left with 155/5

155 divided by 5 is 31

So our final answer is 31

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icang [17]

Answer:

I don't get this.......

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2 years ago

Line segment SU is a diameter at Circle V what is the measure of arc ST?

Inga [223]

The measure of ST based on the diagram regarding the circle is 56°.

<h3>How to calculate the value?</h3>

Fron the information given, angle S is equal to 34°. Also, the triangle is inscribed in a circle.

Therefore, the measure of the angle will be:

= 180 - 90 - 34

= 56°

Therefore, the measure of angle ST is 56°.

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1 year ago

The volume of a spherical sculpture is 256 ft³. Rhianna wants to estimate the surface area of the sculpture. To do the estimate,

Brrunno [24]

Answer:

192 $ft^2$

Step-by-step explanation:

Given that

Volume of spherical sculpture = 256 ft³

$\pi$ is used as 3.

To find:

Surface area of sculpture = ?

Solution:

First of all, let us learn about the formula for Volume and Surface Area of Sphere:

1. $Volume =\frac{4}{3}\pi r^3$

2. $Surface\ Area = 4\pi r^2$

Given volume is 256 ft³.

$256 = \dfrac{4}{3}\pi r^3\\\Rightarrow 256 = \dfrac{4}{3}\times 3 r^3\\\Rightarrow 256 = 4 r^3\\\Rightarrow r^3=64\\\Rightarrow \bold{r = 4\ ft}$

Now, let us put r = 4 in the formula of Surface Area to find the value of Surface Area:

$Surface\ Area = 4\pi 4^2 = 4 \times 3 \times 16 = \bold{192\ ft^2}$

So, approximate surface area of sculpture is 192 $ft^2$ .

3 0

2 years ago

Find the surface area of the cylinder in terms of pi.

Svet_ta [14]

I think the answer is c

4 0

3 years ago

Read 2 more answers

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

2 years ago