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2 years ago

10

(6-2.7h)+(-1.3j-4) ?

2 answers:

SIZIF [17.4K]2 years ago

8 0

Answer:

−2.7h−1.3j+2

Step-by-step explanation:

Murljashka [212]2 years ago

5 0

Answer:

= 2 - 2.7h - 1.3j

Step-by-step explanation:

(6-2.7h)+(-1.3j-4)

we solve the question with the BODMAS

BRACKET

OF

DIVISION

MULTIPLICATION

ADDITION and

SUBTRACTION

(6-2.7h)+(-1.3j-4)

Solve the bracket first

= 6 - 2.7h - 1.3j - 4

Now we collect like terms

= 6 - 4 - 2.7h - 1.3j

= 2 - 2.7h - 1.3j

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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

2 years ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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