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kiruha [24]
3 years ago
7

Find the instantaneous rate of change of f(x)= 4/x+3 at x=5

Mathematics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

(5 , 3.8)

Step-by-step explanation:

Hope this help! :)

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VMariaS [17]

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A.

Step-by-step explanation:

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2627 divided by 22 withR
MrMuchimi

Answer:

119 r 9

Step-by-step explanation:

4 0
1 year ago
Find m&lt; AFE and include working out.
andreyandreev [35.5K]

Answer:

∠ AFE = 80°

Step-by-step explanation:

Since BD is a straight line , then

∠ AFB + ∠ AFE + ∠ EFD = 180°, that is

50° + ∠ AFE + 50° = 180°

100° + ∠ AFE = 180° ( subtract 100° from both sides )

∠ AFE = 80°

3 0
3 years ago
It has been suggested that night shift-workers show more variability in their output levels than day workers. Below, you are giv
bonufazy [111]

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.9

p-value = 0.206

Since the p-value is greater than α therefore, we cannot reject the null hypothesis.

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

Step-by-step explanation:

Let σ₁² denotes the variance of night shift-workers

Let σ₂² denotes the variance of day shift-workers

State the null and alternative hypotheses:

The null hypothesis assumes that the variance of night shift-workers is equal to or less than day-shift workers.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance of night shift-workers is more than day-shift workers.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic or also called F-value is calculated using

Test statistic = Larger sample variance/Smaller sample variance

The larger sample variance is σ₁² = 38

The smaller sample variance is σ₂² = 20

Test statistic = σ₁²/σ₂²

Test statistic = 38/20

Test statistic = 1.9

p-value:

The degree of freedom corresponding to night shift workers is given by

df₁ = n - 1

df₁ = 9 - 1

df₁ = 8

The degree of freedom corresponding to day shift workers is given by

df₂ = n - 1

df₂ = 8 - 1

df₂ = 7

We can find out the p-value using F-table or by using Excel.

Using Excel to find out the p-value,

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.9, 8, 7)

p-value = 0.206

Conclusion:

p-value > α    

0.206 > 0.05   ( α = 1 - 0.95 = 0.05)

Since the p-value is greater than α therefore, we cannot reject the null hypothesis corresponding to a confidence level of 95%

So we can conclude that the night shift workers don't show more variability in their output levels than day workers.

3 0
3 years ago
In ⊙E, what is AC?<br> in need of answers
NNADVOKAT [17]

Answer:

d

Step-by-step explanation:

i looked up the answers

4 0
3 years ago
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