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Aleonysh [2.5K]
3 years ago
9

Hey please help me please help please

Mathematics
1 answer:
Airida [17]3 years ago
4 0

1.

a) metres to centimetres :

multiply length by 100

b) metres to millimetres:

multiply length by 1000

c) kilograms to grams:

multiply the mass value by 1000

d) litres to millilitres :

multiply volume by 1000

2.

a) 3 m = 3× 100 = 300 cm

b) 28 cm = 28 × 10 = 280 mm

c) 2.4 km = 2.4 × 1000

= 24 × 10^-1 × 10^3

= 24 × 10^2 =2400 m

d) 485 mm =485 / 10

= 485 / 10 ^1

= 485 × 10 ^-1

= 48.5 cm

e) 35 cm = 35 / 100

= 35 /10^2

= 35 × 10 ^ -2

= 0.35 m

f) 2.4 m = 2.4 / 1000

= 24 × 10 ^-1 / 10^3

= 24 × 10^-1 × 10 ^-3

= 24 × 10 ^ -4

= 0.0024 km

g) 2495 mm = 2495 /1000

= 2495 /10^ 3

= 2495 × 10 ^-3

=2.495 m

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Answer:

5

Step-by-step explanation:

75/15=5

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Dmitry_Shevchenko [17]

Answer:

x = 5, y = -1/2

Step-by-step explanation:

3.5x - 5y = 20

3x + 4y = 13 multiply by 1.25 so the y's match up (you could match x's too)

3.75x + 5y = 16.25

–––––––––––––––––––––––

3.5x - 5y = 20

3.75x +5y = 16.25

7.25x + 0y = 36.25 you can add or subtract here (on this equation you add)

7.25x=36.25

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3(5) + 4y = 13

15 + 4y = 13

4y = -2

y = -1/2

x = 5, y = -1/2

Then you can plug in both to check your answers.

4 0
3 years ago
Find arc “s”. Theta is 114 degrees
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5 0
3 years ago
An independent-measures research study was used to compare two treatment conditions with n= 12 participants in each treatment. T
Maurinko [17]

Answer:

(a) The data indicate a significant difference between the two treatments.

(b) The data do not indicate a significant difference between the two treatments.

(c) The data indicate a significant difference between the two treatments.

Step-by-step explanation:

Null hypothesis: There is no difference between the two treatments.

Alternate hypothesis: There is a significant difference between the two treatments.

Data given:

M1 = 55

M2 = 52

s1^2 = 8

s2^2 = 4

n1 = 12

n2 = 12

Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(12-1)8 + (12-1)4] ÷ (12+12-2) = 132 ÷ 22 = 6

Test statistic (t) = (M1 - M2) ÷ sqrt [pooled variance (1/n1 + 1/n2)] = (55 - 52) ÷ sqrt[6(1/6 + 1/6)] = 3 ÷ 1.414 = 2.122

Degree of freedom = n1+n2-2 = 12+12-2 = 22

(a) For a two-tailed test with a 0.05 (5%) significance level and 23 degrees of freedom, the critical values are -2.069 and 2.069.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 falls outside the region bounded by the critical values.

(b) For a two-tailed test with a 0.01 (1%) significance level and 23 degrees of freedom, the critical values are -2.807 and 2.807.

Conclusion:

Fail to reject the null hypothesis because the test statistic 2.122 falls within the region bounded by the critical values.

(c) For a one-tailed test with 0.05 (5%) significance level and 23 degrees of freedom, the critical value is 1.714.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 is greater than the critical value 1.714.

6 0
3 years ago
Number three . Show work plz
nordsb [41]
13/15 - 1/3   Find a common denominator 
1/3 x 5= 5/15 
13/15 - 5/15 = 8/15 
8 0
3 years ago
Read 2 more answers
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