Please help me with this question

Dominic is 72 inches tall and casts a 6o inch shadow his son who is standing next to him cast a 50 inch shadow how tall is his s

Lostsunrise [7]

Dominics son is 45inches tall

<u>Step-by-step explanation:</u>

Let his sons height be 'h'

If Dominic is 72 inches and casts a shadow of 60 inches. His son casts a shadow of 50 inches.

Then We can say that,

h = 50 x (72/60)

h = 45 inches

Dominics son is 45inches tall

3 0

3 years ago

Trisha packs x boxes with 12 bottles in each box. The total number of bottles she packs is 216. Which equation and solution repr

LiRa [457]

Answer:

Number of boxes Trisha pack = 18 boxes

Step-by-step explanation:

Given:

Number of boxes Trisha pack = X

Number of bottles in each box = 12

Total number of bottle = 216

Find:

Number of boxes Trisha pack

Computation:

Total number of bottle = Number of boxes Trisha pack x Number of bottles in each box

216 = X × 12

X = 216 / 12

X = 18

Number of boxes Trisha pack = 18 boxes

4 0

3 years ago

Read 2 more answers

8/15-3/15 what’s the answer

kolbaska11 [484]

Answer:

5/15

Step-by-step explanation:

5 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

3 years ago

Some one please help me fast!!

Pie

Answer:

A) 10^23

Step-by-step explanation:

if bases are the same and you're multiplying then:

keep the base and add the exponents

8 0

2 years ago

Please help me with this question​

Please help me with this question