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Pani-rosa [81]
2 years ago
12

How would I do this? Solve for q. –10 + 10q = 9q

Mathematics
1 answer:
DerKrebs [107]2 years ago
4 0

Answer:

q=10

Step-by-step explanation:

-10+10q=9q

-10=-q

q=10

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It represents the ounces of water that the bottle can still hold.
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Please help me figure this out it’s due on 15 minutes
Olin [163]

Answer:

Step-by-step explanation:

1. area of square=10²=100 cm²

area of circle=πr²=π×2.5²=6.25π cm²

reqd. area=100-6.25π ≈80.375 cm²

2.

area of circle=π×2.5²=6.25π≈6.25×3.14≈19.625 cm²

area of rectangle=4×3=12 cm²

reqd. area=19.625-12≈7.625 cm²

3.

area of two circles=2×π×3²=18π≈18×3.14≈56.52 cm²

area of square=12²=144 cm²

reqd. area=144-56.52≈87.48 cm²

4.

area of smaller circle=π×4²=16π cm²

area of bigger circle=π×8²=64π cm²

area of two circles=16π+64π=80π cm²

diameter of bigger circle=4+4+8+8=24 cm

diameter of biggest circle=π×12²=144π

area of shaded region=144π-80π=64π≈200.96 cm²

5 0
3 years ago
Write a letter to the student council that describes
zhenek [66]

Answer:

its b on ege

Step-by-step explanation:

5 0
2 years ago
How to solve a system of three equations using the elimination method
kolbaska11 [484]

Answer:

1. Write all the equations in standard form, leaving out decimals or fractions.

2. Select the variables to be eliminated; And then you take two of these equations, and you eliminate the variables.

3. Select a different set of two equations and eliminate the same variables as in step 2.

4. Solve two equations containing two variables from steps 2 and 3.

5. Substitute the answer to step 4 into any equation that contains the remaining variables.

6. Check the solution with three original equations.

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3 years ago
In order to justify this claim for the Ferengi a sample of 16 coils is taken. For this sample set, the mean warp phase flux is f
Kay [80]

Answer:

The responses to these question can be defined as follows:

Step-by-step explanation:

n = 16\\\\  \bar{x}= 73.7\\\\\sigma = 12\\\\a = 0.05\ or\  a = 0.10\\\\H_{o} \ : \mu = 68\\\\H_{a} \ : \mu \neq 68\\\\a = 0.05\\\\

critical values =\pm t0.025,15 = \pm 2.131\\\\

(n-1) = 15^{\circ}\\\\a = 0.10\\\\

critical values= \pm t0.05,15 = \pm 1.753\\\\

(n-1) = 15^{\circ}\\\\

Testing the statistic values:

t = \frac{x-\mu_{0}}{ \frac{s}{\sqrt{n}}}\\\\

  = \frac{73.7-68}{(\frac{12}{\sqrt{16}})}\\\\\  = \frac{5.7}{(\frac{12}{4})}\\\\  = \frac{5.7}{(3)}\\\\  = 1.9\\

Test statistic ta = -1.90\ lies

The critical values\pm t_{0.05,15} =\pm 1.753

It is in the region of dismissal. We dismiss the 10% significant null hypothesis.

t_a = 1.90 \\\\df = 15\\\\a = 0.05\\\\p-value = 076831\\\\

P - value is greater than the level of significance a= 0.05  

Null hypothesis we don't reject. At a 95% level, the claim is justified.

t_a = 1.90\\\\ df = 15\\\\  a = 0.10\\\\p-value = 076831\\\\

P - value below the meaning level a = 0.10, we reject the hypothesis null. At a level of 90% the claim is not justified.

3 0
2 years ago
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