Please help!!! 2x²-5x-12 (quadratic equations)
2 answers:
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Answer:
The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].
Step-by-step explanation:
We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.
Let X = <u><em>lengths of pregnancies in a small rural village</em></u>
SO, X ~ Normal()
Here, = population mean = 262 days
= standard deviation = 17 days
<u>Now, the 68-95-99.7 rule states that;</u>
- 68% of the data values lies within one standard deviation points.
- 95% of the data values lies within two standard deviation points.
- 99.7% of the data values lies within three standard deviation points.
So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;
[ ,
] = [262 - 17 , 262 + 17]
= [245 days , 279 days]
Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].
Answer:
A
Step-by-step explanation:
Hello There!
First thing we notice is that the y intercept is at (3,0) meaning that they have some sort of fee that has a charge of 3 dollars (represented by the 3)
Because the fee is 3 dollars we can immediately eliminate answers B and C
Now we want to find the charge for each bag of Potpourri (this can be represented by the slope)
Once again we want to choose two points on the line and plug it in into the slope formula
the points are optional but i have chosen the points (1,4.5) and (2,6)
*plug it into the slope formula*
so we can conclude that the charge per bag of potpourri is $1.50 leading us to answer choice A
Answer:
The answers are given in step-by-step explanation
Step-by-step explanation:
In question 4,
Elina walked 12 miles then she walked 0.25 that distance.
We have to find the distance she walked
Distance=12+0.25 of that distance
= 12+0.25(12)
= 12(1+0.25)==15 miles
Hence, A, B, and C are correct
In question 5,
Given circumference is 600 m
we have to find the good approximation area of circle
circumference = 600 m
=300 ⇒ r=95.49
Area of circle is
=
Hence, the good approximation area of circle is
In question 6,
Given d=3t where d represents the distance between the snail and rock in inches and t represents the time in minutes.
Number 3 represents that distance between snail and certain rock is 3 times that of time.
given distance is 9 inches we have to find the time i.e d=3t ⇒ 9=3t ⇒ t=3 minutes.
Hence, will be 3 minutes when snail is 9 inches far from certain rock.
Now, times given 9 minutes we have to find the distance, d=3t=3(9)=27 inches.