Alica has 32 stickers .this is 4 times as many as benita has .how many sticker does benita have?

AfilCa [17]

112
/ \
2 56
/ \
4 14
/ \ / \
2 2 2 7

54
/ \
2 27
/ \
3 9
/ \
3 3

GCF=2

8 0

2 years ago

Emily has 32 grams of a 25% sugar syrup, but this is just too sweet for her! How much pure water should she add to have a 10% sy

GarryVolchara [31]

32 grams = water mass + sugar mass . 100% syrup=75% water+ 25%sugar
original water content= 32*75/100=24 grams
original sugar content= 32*25/100= 8grams

total mass at 10% syrup:
8grams/total mass=10%=0.1
total mass=8/0.1=80grams
total mass of water=80grams- 8grams of sugar content= 72 grams
total mass of water added=72grams- 24grams original water content= 48grams.

answer: 48grams

3 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:

A)

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

C)

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

The diagram shows a solid shape.

Stolb23 [73]

Answer:

$756\pi$

Solution,

Volume of cone=270 pi cm^3

$\frac{1}{3} \pi \: {r}^{2} = 270\pi \\ or \: {r}^{2} \times 10 = 270 \times 3 \\ or \: {r}^{2} \times 10 = 810 \\ or \: {r}^{2} = \frac{810}{10} \\ or \: {r}^{2} = 81 \\ or \: r = \sqrt{81} \\ or \: r = \sqrt{ {9}^{2} } \\ r = 9 \: cm$