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andrezito [222]
3 years ago
14

Translate the sentence into an inequality. Twice the difference of a number and 8 is at least 26 . Use the variable w for the un

known number.
Mathematics
1 answer:
Olin [163]3 years ago
7 0
"At least" refers to greater than or equal to (≥).
This is because "at least" means:
- it can't be less than
- it can be equal to
- it can be greater than
This is basically like saying no less than.
If you must have at least 10 oranges for some recipe, then you must have 10 oranges, but you can also have more.

"Twice the difference of a number and 8"
This means that 2 times a number minus 8.
This can be written as 2(w - 8), since w is unknown and it's the difference between w and 8.
The 2 is being multiplied by this because it's 2 times this difference.

"At least 26"
This means ≥ 26.

Now for the actual inequality.
"Twice the difference of a number and 8"
translates to 2(w - 8)
"is at least 26"
translates to ≥ 26
Now put them together.
2(w - 8) ≥ 26

So the answer is 2(w - 8) ≥ 26.

Hope this helps! :)
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Write a rule for the nth term of the arithmetic sequence 3, 11, 19, 27, 35... what is a20?
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3, 11, 19, 27, 35...

a = 3
d = 11-3 = 8
a20 = a + 19d = 3 + 19*8 = 3 + 132 = 135

HOPE IT HELPS!
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3 years ago
The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp
vovikov84 [41]

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x 10^{-5} .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, \sigma = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm   and Sample size, n = 15

(a) Let Null Hypothesis, H_0 : Mean Diameter, \mu = 8.2500 cm

 Alternate Hypothesis, H_1 : Mean Diameter,\mu \neq 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

                           \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } follows Standard Normal distribution

After putting each value, Test Statistics = \frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } } = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172  and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x 10^{-5} .

(c) For constructing Two-sided confidence interval we know that:

    Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 <  \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.96) = 0.95

P(-1.96\frac{\sigma}{\sqrt{n} } < Xbar - \mu < 1.96\frac{\sigma}{\sqrt{n} } ) = 0.95

P(-Xbar-1.96\frac{\sigma}{\sqrt{n} } < -\mu < 1.96\frac{\sigma}{\sqrt{n} }-Xbar ) = 0.95

P(Xbar-1.96\frac{\sigma}{\sqrt{n} } < \mu < Xbar+1.96\frac{\sigma}{\sqrt{n} }) = 0.95

So, 95% confidence interval for \mu = [Xbar-1.96\frac{\sigma}{\sqrt{n} } , Xbar+1.96\frac{\sigma}{\sqrt{n} }]

                                                        = [8.2535-1.96\frac{0.0020}{\sqrt{15} } , 8.2535+1.96\frac{0.0020}{\sqrt{15} }]

                                                        = [8.2525 , 8.2545]

Here \mu = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

           =  [8.2525 , 8.2545] .

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