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BARSIC [14]
4 years ago
8

What is the volume? 17 cm 13 cm 5 cm It’s a triangular prism.

Mathematics
1 answer:
aksik [14]4 years ago
3 0

Answer:

1105 cubic centimetre

Step-by-step explanation:

volume of rectangular prism

= 17 \times 13 \times 5 \\  = 1105 \:  {cm}^{3}

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Which table shows the same relationship as y = -x^2 + 3x?
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Answer:

H

Step-by-step explanation:

given y = -x² + 3x.

let f(x) be y.

f(-2) = -(-2)² + 3(-2) = -10

f(-1) = -(-1)² + 3(-1) = -4

f(0) = -(0)² + 3(0) = 0

f(1) = -(1)² + 3(1) = 2

f(2) = -(2)² + 3(2) = 2

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3 years ago
Help me please asap ‘,:’
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0.75

Step-by-step explanation:

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Type the correct answer in the box. Use the information from the image to complete the table then determine the total number of
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When the integer n is squared, the result is less than 150. What is the sum of all possible values of n?
Rom4ik [11]
Answer = 0 

n can have possible values from (-12)  to 12
-12 <= n < 12

(-12) + (-11)+(-10)......+10+11+12=0
5 0
4 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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