Answer:
y= -3
Step-by-step explanation:
Since there is no slope, there is no mx in y=mx+b. Because the line has to go through point (6,-3), the y-intercept is -3.
Answer:
112.41
Step-by-step explanation:
9 percent of 1,249 is 112.41
Answer:
<em>6 cm </em>
Step-by-step explanation:
<u><em>I don't know may I use the trigonometry functions, therefor I'm going to use Pythagorean theorem and property of 30°-60°-90° triangle.</em></u>
m∠ACB = 120° ⇒ m∠ACD = 60° ⇒ m∠DAC = 30°
DC is opposite to 30° angle ⇒ AC = CB = 8 cm
From ΔADC: AD² = 64 - 16 = 48 ; AD = 4√3 cm
ΔACB is isosceles ⇒ m∠CAB = m∠CBA = 30°
m∠DAH = m∠DAC + m∠CBA = 2(30°) = 60° ⇒ m∠ADH = 30° ⇒ AH = 4√3 ÷ 2 = 2√3 cm
From ΔADH: <em>DH</em> =
= √36 = <em>6 cm</em>
Using the frequency table and the probability concept, it is found that there is a 0.5581 = 55.81% probability that the waiting time is at least 12 minutes or between 8 and 15 minutes.
- A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.
- In this problem, the frequency table gives these numbers of outcomes.
At least 12 minutes or between 8 and 15 minutes is equivalent to <u>at least 8 minutes,</u> thus at least 8 minutes is our desired outcome.
- There are 9 + 10 + 12 + 4 + 4 + 2 + 2 = 43 customers.
- Of those, 24 spent at least 8 minutes.
Thus, the probability is:

0.5581 = 55.81% probability that the waiting time is at least 12 minutes or between 8 and 15 minutes.
A similar problem is given at brainly.com/question/15536019
This question is incomplete, the complete question is;
A certain organization reported the following scores for two parts of the scholastic Aptitude test ( SAT)
Evidence-based Reading and writing : 533
Mathematics : 527
Assume the population standard deviation for each part is σ = 100.
What is the probability a sample of 66 test takers will provide a sample mean test score within 10 points of the population mean of 533 on the Evidence-based Reading and Writing part of the test?
Answer: the required probability is 0.582
Step-by-step explanation:
Given that;
Population mean = 533
sample size n = 66
population standard deviation σ = 100
σ of x bar = 100/√66 = 12.3091
Normal distribution with mean 533 and SD of 12.3091
P( 523 <x< 543 )
Z = 10 / 12.3091
Z = 0.8124, -0.8124
P( z < 0 0.8124) - P( z < -0.8124) { from table}
⇒ 0.7910 - 0.2090
= 0.582
Therefore, the required probability is 0.582