MATHEMATICS HIGH SCHOOL

The diameter of a circle is a chord that _______ passes through the center of the circle. A. always
B. sometimes
C. never

Answers

Answer 1
Answer: The answer is A. Always.
The diameter of a circle is a line that goes from one side to the other hence passing through the center!
Hope this helped!
Answer 2
Answer:

Answer:

The answer is A. always

Step-by-step explanation:

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Katie types 600 words in 5 minutes. She compared her typing speed to the record typing speed for her school shown in the table.Choose the answer that correctly compares Katie’s typing speed to the record typing speed for her school.

Question options:

A.) Katie types slower than the school record speed.
B.) Katie types faster than the school record speed.
C.) Katie’s typing rate is the same as the record speed for her school.

Answers

Answer:

a

Step-by-step explanation:

since the school recorded is in two minutes, they wrote 420 words, and Katie in 5 minutes wrote 600 words, if you put the average to the person who wrote 420 words in 2 minutes to-

in every too minutes they write 420 words.

so in four minutes they would of typed 840 words, which is more words than Katie in less minutes.

-3 progression 8 common difference what is the 28th term

Answers

the answer for this problem would be 213

Solve and then use estimation to make sure your answer is reasonable. Kyle can read 22 pages of his book in an hour. Kyle read for 114.5 hours last summer.

How many pages did Kyle read last summer?

Answers

114.5 hours = 6870 minutes
∴ 6870 ÷ 22 = 312.2727273

What is the solution set for the inequality |-3-5x|-8<-2

Answers

|-3-5x|-8<-2\n|-3-5x|<6\n-3-5x<6 \wedge -3-5x>-6\n-5x<9 \wedge -5x>-3\nx>-(9)/(5) \wedge x<(3)/(5)\nx\in\left(-(9)/(5),(3)/(5)\right)
The hard way:

\left| -3-5x \right| -8<-2\n \n \left| -3-5x \right| <6\n \n { \left( -3-5x \right) }^( 2 )<{ 6 }^( 2 )\n \n \left( -3-5x \right) \left( -3-5x \right) <36

\n \n 9+15x+15x+25{ x }^( 2 )<36\n \n 25{ x }^( 2 )+30x+9<36\n \n 25{ x }^( 2 )+30x-27<0\n \n Say\quad f\left( x \right) =25{ x }^( 2 )+30x-27,\n \n and\quad that\quad f\left( x \right) =0

\n \n 25{ x }^( 2 )+30x-27=0\n \n 25{ x }^( 2 )+30x=27\n \n { x }^( 2 )+\frac { 30 }{ 25 } x=\frac { 27 }{ 25 } \n \n { x }^( 2 )+\frac { 6 }{ 5 } x=\frac { 27 }{ 25 } \n \n { \left( x+\frac { 3 }{ 5 } \right) }^( 2 )-{ \left( \frac { 3 }{ 5 } \right) }^( 2 )=\frac { 27 }{ 25 }

\n \n { \left( x+\frac { 3 }{ 5 } \right) }^( 2 )=\frac { 36 }{ 25 } \n \n x+\frac { 3 }{ 5 } =\pm \frac { 6 }{ 5 } \n \n x=-\frac { 3 }{ 5 } \pm \frac { 6 }{ 5 }

Therefore:\n \n x=\frac { 3 }{ 5 } \quad and\quad x=-\frac { 9 }{ 5 } \quad when\quad f\left( x \right) =0\n \n Now:\n \n f\left( x \right) <0,\n \n

When:

-9/5<x<3/5

Given f(t)=t to the power of 4 +t² -5, find the value of f(b) - f(-b)

Answers

f(t)=t^4+t^2-5\n \n f(b)-f(-b)=b^4+b^2-5-[(-b)^4+(-b)^2-5]\n \n f(b)-f(-b)=b^4+b^2-5-b^4-b^2+5\n \n \boxed{f(b)-f(-b)=0}
f(b)-f(-b)=b^4+b^2-5-((-b)^4+(-b)^2-5)\n f(b)-f(-b)=b^4+b^2-5-(b^4+b^2-5)\n f(b)-f(-b)=b^4+b^2-5-b^4-b^2+5\n f(b)-f(-b)=0

On discovering that her family had a 70% risk of heart attack, Erin took a treadmill test to check her own potential of having a heart attack. The doctors told her that the reliability of the stress test is 67%. The test predicted that Erin will not have a heart attack. What is the probability after the test was taken that she will have a heart attack?A. 0.4051

B. 0.5010

C. 0.4653

D. 0.6632

Answers

On discovering that her family had a 70% risk of heart attack, Erin took a treadmill test to check her own potential of having a heart attack. The doctorstold her that the reliability of the stress test is 67%. The test predicted that Erin will not have a heart attack. The probability after the test was taken that she will have a heart attack is "B. 0.501."

Answer:

The probability that she will not have a heart attack and the test predicts that she will is 0.4653 or 46.53%

Step-by-step explanation:

Hint- This a conditional probability problem where Bayes theorem should be applied.

Applying Bayes theorem,

P(\text{No heart attack}\ |\ \text{Correctly tested})=

\frac{P(\text{Correctly tested}\ |\ \text{No heart attack})\cdot P(\text{No heart attack})}{P(\text{Correctly tested})}

P(\text{Correctly\ tested}\ |\ \text{No\ heart\ attack})=67\%=0.67

P(\text{No\ heart\ attack})=1-P(\text{heart\ attack})=1-0.7=0.3

P(\text{Correctly\ tested})=[P(\text{No\ heart\ attack})* P(\text{Correctly\ tested})]+[P(\text{Heart\ attack})* (\text{Incorrectly\ tested})]

=[0.3* 0.67]+[0.7* 0.33]=0.432

Putting the values,

P(\text{No\ heart\ attack}\ |\ \text{Correctly\ tested})=(0.67* 0.3)/(0.432) =0.4653

∴ There is a probability of 0.4653 or 46.53% chance that she will not have a heart attack even though the test predicts that she will.
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