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EastWind [94]
3 years ago
10

Math help me pls now now

Mathematics
2 answers:
emmainna [20.7K]3 years ago
8 0

Answer: 4 1/12

Step-by-step explanation:

Hope this helps!

JulijaS [17]3 years ago
6 0

Answer:

14/1

Step-by-step explanation:

16/3 –5/4

(16*4–5*3)/12

49/12

1 4/1

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Question 9 (1 point)
Romashka [77]

Answer:

Step-by-step explanation:

League A                   League B

151.12                            163.25

148                                157

26.83                             24.93

29                                  136

136                                145

167                                178

207                               256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(82.5-1.5\times 76.56,159.06+1.5\times 76.56)

(-32.34,273.9)

So, There is no outlier

Maximum = 207

2)

League B in ascending order :

24.93,136,145,157,163.25,178,256

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}==\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

(140.5-1.5\times 30.125,170.625+1.5\times 30.125)

(95.3125,215.8125)

24.93 and 256 are outliers  

Maximum = 256

5 0
3 years ago
Xhoose the word incorectlycomplete the sntence
Semenov [28]
This question needs more information
7 0
3 years ago
What is the distance between points (10, 7)(10, 7) and (2, 7)(2, 7) on a coordinate plane?
lawyer [7]
I am sorry i have no idea. :( But i can ask my brother who is in a higher grade than i. i will ask him when he gets home for you. :)
5 0
4 years ago
The mean time it takes to walk to the bus stop is 8 minutes (with a standard deviation of 2 minutes) and the mean time it takes
kifflom [539]

We have been given for a normal distribution the mean time it takes to walk to the bus stop is 8 minutes with a standard deviation of 2 minutes. And the mean time it takes for the bus to get to school is 20 minutes with a standard deviation of 4 minutes.

(a) Average time that it would take reach school can be obtained by adding the average times.

8+20 = 28 minutes.

(b) Standard deviation of the trip to school can be found as:

\sigma =\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=4.47

Therefore, standard deviation of the entire trip is 4.47 minutes.

(c) Let us first find z score corresponding to 30 minutes.z=\frac{x-\mu }{\sigma }=\frac{30-28}{4.47}=0.447

We need to find the probability such that P(x>30)=P(z>0.447)=0.67

Therefore, the required probability is 0.67.

(d) If average time to walk to school is 10 minutes, then overall average time for the trip will be 10+20 = 30 minutes.

(e) Standard deviation won't change it will remain 4.47

(f) The new probability will be:

z=\frac{x-\mu }{\sigma }=\frac{30-90}{4.47}=0

P(x>30)=P(z>0)=0.5

Therefore, probability will be 0.50.


6 0
3 years ago
Read 2 more answers
Kit bought a soft drink and a sandwich for $9.00. What was the price of each if the sandwich cost 3.5 times much as the soft dri
astra-53 [7]

Let the cost of the soft drink = C   then the price of the sandwich was 3.5 times as much = 3.5C

 

And

 

C + 3.5C  = Total Cost

 

C + 3.5C  =   9

 

4.5C = 9    divide both sides by 4.5

 

C = 9/4.5  =  $2.00   for the soft drink

 

And  3.5C = 3.5(2.00)  = $7.00   cost of the sandwich

8 0
3 years ago
Read 2 more answers
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