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vredina [299]
3 years ago
10

What is the answer can you help me plz

Mathematics
1 answer:
erastova [34]3 years ago
6 0

Answer:

\frac{1}{5^{7}}

Step-by-step explanation:

\frac{5^{-4}}{5^{3}} = \frac{1}{5^{7}}

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For 33 hrs of work you are paid 252.45 how much would you receive for 36 hours
poizon [28]
For 33 Hours, It is = $252.45

Now, for 1 Hour, It is 252.45 / 33 = 7.65

Now, For 36 hours, It would be: 7.65 * 36 = 275.4

In short, Your Answer would be: $275.4

Hope this helps!
3 0
3 years ago
Find the volume of a cone
lianna [129]

Answer:

96π or 301.59 in³

Step-by-step explanation:

V(cone) = 1/3πr²h

= 1/3·π·6²·8

= 96π or 301.59

6 0
3 years ago
Read 2 more answers
Someone help me out with this question please
storchak [24]

Answer: 9.1

Step-by-step explanation:

Add all of the numbers up and divide by 8

6 0
3 years ago
Read 2 more answers
PLEASE HELP
EleoNora [17]

Answer:

35N

Step-by-step explanation:

Find the constant multiplied to the mass:

8 * k = 40

Divided by 8

k = 5

Find the weight of a 7 kg mass

7 * 5 = 35N

5 0
3 years ago
Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answ
gulaghasi [49]

Answer:

Part a ) The degrees of freedom for the given two sample non-pooled t test is 24

Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

Δf  = 24.3223 ≈ 24 (down to the nearest whole number)

b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

6 0
3 years ago
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