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3 years ago

What is the length of chord cd in 0 below

Mathematics

2 answers:

nikdorinn [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Because it is at same distance as other one from center so they have same lengths

LenKa [72]3 years ago

4 0

Answer:

I believe 5 units

Step-by-step explanation: I believe it is five units because the other one Chord AB is also 5 units. Plz mark brainliest if correct

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What is the answer for -7 = 6-1/5x

oksano4ka [1.4K]

Answer:

x = 65

Step-by-step explanation:

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3 years ago

49 out of the 98 teachers in a school staff meeting were first-year teachers. What percentage of the teachers in the meeting wer

Evgesh-ka [11]

Answer:

50% of the teachers in the meeting were first-year teachers.

Step-by-step explanation:

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2 years ago

I need answers I’ll cash app 10 dollars

posledela

1. 0 > 3x - 3 - 6

0 > 3x -9

9 > 3x

3 > x

It would be a closed circle starting from 3 going to the right.

2. 4x + 1 - 1 >/ -8

4x >/ -8

x >/ -2

It would be an open circle starting from -2 going to the right.

3. -1 </ 2n + 4 - 5

-1 </ 2n -1

0 </ 2n

Undefined.

4. -6 > 5n + 5 + 4

-6 > 5n + 9

-15 > 5n

-3 > n

It would be a closed circle going to the right

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3 years ago

a meter is a unit of length approximately equal to 39.37 inches. if someone is 1.98 meters tall what is their approximate height

galina1969 [7]

78 inches is the correct answer.

5 0

3 years ago

Read 2 more answers

A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the prob

UkoKoshka [18]

Answer:

The probability that 8 mice are required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates mice, one at a time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

$P(X=r)=^nC_r p^rq^{n-r}$

Where, p is the probability of success $p=\frac{2}{7}$

q is the probability of failure q=1-p, $q=1-\frac{2}{7}=\frac{5}{7}$

n is total number of trials n=8

r=3

Substitute the values,

$P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}$

$P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}$

$P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}$

$P(X=3)=0.2428$

Therefore, the probability that 8 mice are required is 0.2428.

5 0

3 years ago