Answer:
a) 0.283 or 28.3%
b) 0.130 or 13%
c) 0.4 or 40%
d) 30.6 mm
Step-by-step explanation:
z-score of a single left atrial diameter value of healthy children can be calculated as:
z=
where
- X is the left atrial diameter value we are looking for its z-score
- M is the mean left atrial diameter of healthy children (26.7 mm)
- s is the standard deviation (4.7 mm)
Then
a) proportion of healthy children who have left atrial diameters less than 24 mm
=P(z<z*) where z* is the z-score of 24 mm
z*=
≈ −0.574
And P(z<−0.574)=0.283
b) proportion of healthy children who have left atrial diameters greater than 32 mm
= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm
z*=
≈ 1.128
1-P(z<1.128)=0.8703=0.130
c) proportion of healthy children have left atrial diameters between 25 and 30 mm
=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm
z(30)=
≈ 0.702
z(25)=
≈ −0.362
P(z<0.702)=0.7587
P(z<−0.362)=0.3587
Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4
d) to find the value for which only about 20% have a larger left atrial diameter, we assume
P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.
Then P(z<z*)=0.8 and z*=0.84. That is
0.84=
solving this equation for X we get X=30.648
The correct answer is A. Jimmy is running late, so he starts to run to school but needs to take breaks.
Explanation:
The graph shows the distance in axis y and the time in axis x. Additionally, the graph presents different sections from A to E. In this, the sections A, C, and E show an increase in the distance from home, this implies there was movement. Moreover, the speed (distance traveled in time) is higher in sections C and E than in A because the distance increases in a shorter time. Also, in sections D and B there is no movement as time continues but the distance is the same. In this context, the description that best matches the graph is "Jimmy is running late, so he starts to run to school but needs to take breaks" because this is the only option that includes the breaks or lack of movement in sections B and D. Also, the changes in speed are likely to occur in this scenario.
<u>Answer:</u>
He must sell 190 posters make a profit of $300.
<u>Explanation:
</u>
Given:
Base fee= $270
Fee for supplies= $2
Cost of poster=$5
To find:
The number of posters to be sold to get a profit of $300=?
Solution:
Let the number of posters sold be n.
Now, we know that,
profit = selling price – cost price
$ 300 = $ 5 for each poster – ( $ 270 base fee + $ 2 for each poster)
300 = 5 x n – (270 + 2 x n)
300 = 5n – 270 – 2n
5n – 2n = 300 + 270
3n = 570
n = 190
Answer:
If it's of which decimal than the answer will be 0.09
= 7.
Ratio means multiply/divide.
To get the next number, multiply number by common ratio.
To get a previous number, divide by common ratio.
Since we want the next number...
2nd term: 7 x -2 = -14
3rd term: -14 x -2 = 28
4th term: 28 x -2 = -56
5th term: -56 x -2 = 112
Keep going until 9th term!
