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2 years ago

12

If you are constructing a 95% confidence interval for a normally distributed population when your sample size is 10, what value

should you use for tα/2? (round to two decimal places)

1 answer:

Nuetrik [128]2 years ago

3 0

This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.

If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"

I show an example below of what I mean.

In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26

<h3>Answer: 2.26</h3>

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Rudik [331]

Answer: see photo

Step-by-step explanation:

5 0

1 year ago

Given that Ray E B bisects ∠CEA, which statements must be true? Select three options. m∠CEA = 90° m∠CEF = m∠CEA + m∠BEF m∠CEB =

FromTheMoon [43]

Answer:

Here is the question attached with.

$m\angle CEA =90 \ (deg)$

$m\angle BEF=135\ (deg)$

$\angle CEF$ is a straight line.

$\angle AEF$ is a right angled triangle.

Options $1,4,5,6$ are correct answers.

Step-by-step explanation:

⇒As $\ ray\ AE$ is ⊥ $FEC$ so it will forms right angled triangle then $m\angle CEA =90\ (deg)$ .

⇒Measure of $\angle BEF =135\ (deg)$ as $\angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg)$ as $\angle AEB$ is the bisector of $\angle AEC$ ,meaning that $\angle AEB$ is half of $\angle AEC$ so $\angle AEB = 45\ (deg)$ .

⇒ $\angle CEF$ is a straight line as the angles measure over it is $180\ (deg)$ .

⇒Measure of $\angle AEF = 90\ (deg)$ from linear pair concept.

As $\angle CEA + \angle AEF = 180\ (deg)$ ,plugging the values of $m\angle CEA =90\ (deg)$ we have $\angle AEF = 90\ (deg)$ .

The other two options are false as:

$m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225$

it is exceeding $180\ (deg)$ whereas $\angle CEF$ is a

straight line.

And $m\angle CEB=2(m\angle CEA)$ is not true.

As $\angle CEA = 90\ (deg)$ and $\angle CEB=45\ (deg)$

So we have total $4$ answers.

The correct options are $1,4,5,6$ .

5 0

2 years ago

Read 2 more answers

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

A.

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

B.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$

Then

$\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.$

Option B is false.

3.

$\sin C = \dfrac{\cos A}{\tan C}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

Now

$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

D.

$\cos A=\tan C.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

As you can see $\cos A\neq \tan C$ and option D is not true.

E.

$\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.$

By the definition,

$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

This option is false.

8 0

2 years ago

Read 2 more answers

Triangle A and triangle B are drawn on the grid. Describe fully the single transformation which maps triangle A onto triangle B.

madam [21]

Answer:

Step-by-step explanation:

Triangle A, transforms into a smaler size, and goes into full shape. Triangle B, goes into the negative numbers.

7 0

1 year ago

Find FL if H is the circumcenter of EFG,

Zanzabum

The artistic crop isn't helpful; it cuts off some vertex names.

The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so

EH² = HL² + EL²

EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339

Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.

Answer: 4.25

8 0

2 years ago