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1 year ago

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From 2000-2003, students were tested by the state in four major subject areas of math, science, English and social studies. The

chart below displays average scores for all students at cherrybrook high school. In which subject did students score best on average across these four years.

1 answer:

Verizon [17]1 year ago

8 0

No way to tell . . . . . we can't see the chart below.
It must be WAY down there where the sun don't shine.

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Please help me answer this ASAP! The two-way table shows the number of ninth and tenth graders who prefer going to sporting even

Elis [28]

<h2>Answer</h2>

0.43

<h2>Explanation</h2>

Remember that $RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}$

Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so $Frecuency=6$ . Now, the sum of all frequencies will be the sum of all the 10th graders, so $SumOfAllFrecuencies=6+8=14$ . Let's replace the values:

$RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}$

$RelativeFrecuency=\frac{6}{14}$

$RelativeFrecuency=0.4285$

And rounded to the nearest hundredth:

$RelativeFrecuency=0.43$

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1 year ago

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

1 year ago

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If c = 10d, what percent of 3c is 3d?

son4ous [18]

Answer:

1000%

Step-by-step explanation:

c=10d,

3c will become 3*10d=30d

d is 10 times to c which converts to 1000%

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1 year ago

297,060 / 0.0004839 =<br><br> Answer in Scientific Notation it Standard Notation

Valentin [98]

Answer:

6.1389 x 10^8

Step-by-step explanation:

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1 year ago

The endpoints of line AB are A(2, 2) and B(3, 8). Line AB is dilated by a scale factor of 3.5 with the origin as the center of d

Musya8 [376]

A` ( 7, 7 )
B ` ( 10.5, 28 )
The slope: m = (28-7) / ( 10.5 - 7 ) = 21 / 3.5 = 6
d ( A` B `) = √ ( 10.5 - 7 )² + ( 28 - 7 )² = √ 3.5² + 21² =
= √ 12.25 + 441 = √ 12.25 ( 1 + 36 ) = 3.5 √37 ( or 3.5 * (37) ^(1/2))
Answer:
C ) m = 6, A`B` = 3.5√37

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1 year ago

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