Colin invests £300 into his bank account.

Please Help! I will give you the brainiest and a lot of points!

slamgirl [31]

#19

$\\ \sf\longmapsto \dfrac{7}{5}$

#20

$\\ \sf\longmapsto \dfrac{9}{5}$

#21

$\\ \sf\longmapsto \dfrac{11}{7}$

#22

$\\ \sf\longmapsto \dfrac{12}{7}$

#23

$\\ \sf\longmapsto \dfrac{2}{2}=1$

#24

$\\ \sf\longmapsto \dfrac{3}{2}$

#25

$\\ \sf\longmapsto \dfrac{5}{2}$

#26

$\\ \sf\longmapsto \dfrac{6}{2}=3$

#27

$\\ \sf\longmapsto \dfrac{8}{2}=4$

#28

$\\ \sf\longmapsto \dfrac{9}{2}$

#29

$\\ \sf\longmapsto \dfrac{13}{4}$

#30

$\\ \sf\longmapsto \dfrac{19}{4}$

#31

$\\ \sf\longmapsto \dfrac{37}{6}$

#32

$\\ \sf\longmapsto \dfrac{25}{6}$

4 0

3 years ago

Read 2 more answers

Please someone help with this question asap

olga nikolaevna [1]

Answer:

-19

Step-by-step explanation:

f(x) is the same as solving for the variable y

So y= -4(7)+9

f(7)= -28+9 = -19

6 0

3 years ago

A como of rope 603 feet long is cut into 25 equal pieces. What is the length of each piece? Express your answer as a mixed numbe

Anna007 [38]

603/25=24 and 3/25 (mixed number)
24.12 (decimal)

3 0

4 years ago

Please help me it's due at 6pm and i promise the give brainlist

Mrac [35]

Answer:

i would say it is b because its more likely to happen

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The following data are from an experiment designed to investigate the perception of corporate ethical values among individuals w

alukav5142 [94]

Answer:

Part 1

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. the F value lies in the rejection region > 3.6823

3. The value of the test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

Since p-value < α, H0 is rejected.

6. Between x`2 and x`3

7. Fisher's Least Significant Difference value almost 0.869

8.There is a significant difference between the means

Step-by-step explanation:

Summary of Data

Treatments

1 2 3 Total

n 6 6 6 18

∑x 42 57 30 129

Mean 7 9.5 5 7.167

∑x2 298 543 152 993

Sd.D 0.8944 0.5477 0.6325 2.0073

ANOVA Table

Source SS df MS

Between-treatments 61 2 30.5 F = 61

Error 7.5 15 0.5

Total 6 8.5 17

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. Using alpha= 0.05 the F value lies in the rejection region i.e F > 3.6823

x1` -x2`= 7-9.5= -2.5 Not significant as difference <3.68

x1`- x3`= 7-5= 2 Not significant as difference <3.68

x2` -x3`= 9.5-5= 4.5 Significant as difference > 3.68

3. The value of the F test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

Since p-value < α, H0 is rejected.

6. Using alpha= .05, differences occurs between x2` and x3` as their difference is greater than 3.68

7. Fisher's Least Significant Difference value almost 0.869

Least Significant Difference= t( 0.025,15) √2s²/r s²= 0.50 r= 6 =n1=n2=n3

Least Significant Difference= 2.13 √ 2*0.50/ 6

=0.869

8.There is a significant difference between the means

x1` -x2`= 7-9.5= -2.5 Significant as difference > Least Significant Difference

x1`- x3`= 7-5= 2 Significant as difference > Least Significant Difference

x2` -x3`= 9.5-5= 4.5 Significant as difference > Least Significant Difference

6 0

3 years ago