All categories

3 years ago

-6.25x>-38 3/4 solve the inequality

Mathematics

1 answer:

Nostrana [21]3 years ago

8 0

Answer:

Step-by-step explanation:

-6.25x > -38 3/4

-6.25x > -155/4

Multiply by (-4)

25x > 155

divide by 25,

x > 155/25

x > 31/5

x > 6.2

You might be interested in

(a^2-4)(9-b^2)+24ab

Sholpan [36]

<span> -a2b2 + 9a2 + 24ab + 4b2 - 36 </span><span>hope it helps

</span>

7 0

3 years ago

Which of the following are true statements.

Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

$c^{2} =a^{2} +b^{2}$ ----------(1)

Let $c=41$ and $a = 7$ and $b=40$

Put all the value in equation 1.

$41^{2} =7^{2} +40^{2}$

$1681=49+1600$

$1681=1649$

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let $c=20$ and $a = 12$ and $b=16$

$20^{2} =12^{2} +16^{2}$

$400=144+256$

$400=400$

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0

3 years ago

Bas_tet [7]

Answer:

a) We reject H₀ We find difference between the mean of the two groups

b) we accept H₀ : There is no statistical difference ( in the scale of pain) when the difference in samples means is 1

Step-by-step explanation:

Group 1: Only with educational material

Sample size n₁ = 42

Sample mean x₁ = 3,2

Sample standard deviation s₁ = 2,3

Group 2: With exercises with the treatment of individuals

Sample size n₂ = 42

Sample mean x₂ = 5,2

Sample standard deviation s₂ = 2,3

Test Hypothesis

Null Hypothesis H₀ x₂ - x₁ = 0 or x₁ = x₂

Alternative Hypothesis Hₐ x₂ - x ₁ > 0 or x₂ > x₁

Significance level α = 0,01

Sample sizes are n₁ = n₂ > 30

We use z-test

for α = 0,01 z(c) ≈ 2,32

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √ (2,3)²/42 + (2,3)²/42

z(s) = ( 5,2 - 3,2 )/ √0,2519

z(s) = 2 / 0,5

z(s) = 4

Comparing z(s) and z(c)

z(s) > z(c)

z(s) is in the rejection region. We reject H₀ the true average pain level for the control condition exceeds that for the treatment condition.

b) Does the true average pain level for the control condition exceed that for the treatment condition by more than 1.

In this case

z( s) = x₂ - x₁ / 0,5

z(s) = 1 /0,5

z(s) = 2

Comparing z(s) and z(c) now

z(s) < z(c) 2 < 2,32

And z(s) is in the acceptance region ( for the same significance level) and we should accept H₀ equivalent to say that the two groups statistical have the same mean

5 0

3 years ago

HELP... which equation is graphed here?

Studentka2010 [4]

i believe the answer would be A.

in slope-intercept form, it's y = -3/2x - 1

3 0

3 years ago

Read 2 more answers

1). ppllllsssss awwnnsseeer thisss

Sophie [7]

Answer: