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Oxana [17]
3 years ago
5

Which inequality is true?

Mathematics
2 answers:
Paul [167]3 years ago
5 0

Answer: 2. Is true

Step-by-step explanation: (second option is true)

mrs_skeptik [129]3 years ago
4 0

Answer:

\frac{3}{4} > \frac{2}{3}

Step-by-step explanation:

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Are 30:12 and 40:16 equivelant if noy tell why
mihalych1998 [28]
They are equivalent

30:12 => 5:2
40:16 => 5:2

hope this helsp
4 0
3 years ago
HELP ME <br> solve for the right triangle
VikaD [51]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle

cos41° = \frac{adjacent}{hypotenuse} = \frac{VW}{VX} = \frac{7}{VX}

Multiply both sides by VX

VX × cos41° = 7 ( divide both sides by cos41° )

VX = \frac{7}{cos41} ≈ 9.3

-----------------------------------------------------------------------

tan41° = \frac{opposite}{adjacent} = \frac{WX}{VW} = \frac{WX}{7}

Multiply both sides by 7

7 × tan41° = WX, thus

WX ≈ 6.1

-------------------------------------------------------------------------

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the given angles from 180° for ∠ X

∠ X = 180° - (90 + 41)° = 180° - 131° = 49°

3 0
2 years ago
Answer quick plsssssssssss thx
pentagon [3]

Answer:

I think Substitution

Step-by-step explanation:

Because x is already solved for in the second equation.

6 0
2 years ago
Perform the indicated operation 7 1/8 times 5 2/3
alukav5142 [94]
35/12. When you multiply the operation this is the answer you get.
3 0
2 years ago
Read 2 more answers
A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
lina2011 [118]

Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0
3 years ago
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