Answer: The difference between M and mu is due to the sampling error is <u>-6.59
.</u>
Step-by-step explanation:
As per given:
in population, the average number of minutes spent in active play on weekends is μ = 65.87
In sample, the average number of minutes the children spend in active play on weekends is M = 59.28
Now, the difference between M and μ is due to the sampling error is M- μ
= 59.28 - 65.87
= -6.59
So, the difference between M and mu is due to the sampling error is <u>-6.59
.</u>
Step-by-step explanation:
there are 2 similar triangles : ABE and DCE
that means they have the same angles, and the scaling factor from one triangle to the other is the same for every side.
and that means that
DE/AE = EC/BE (= DC/AB)
we know that
AE = AD + DE
BE = BC + EC
a.
so, we have actually
DE/(AD+DE) = EC/(BC+EC)
DE/(10+DE) = 8/(2+8) = 8/10 = 4/5
DE = 4(10+DE)/5
5DE = 4(10+DE) = 40 + 4DE
DE = 40 cm
b.
AD/DE = 3/5
BC/EC must be 3/5 too.
15/EC = 3/5
15 = 3EC/5
75 = 3EC
EC = 25 cm
0.5625 will be the answer to 9 divided by 16
Answer:
Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.
Step-by-step explanation:
Given:
First 4 test scores = 87%, 92%, 76%,89%
Average targeted = 80%
We need to find the minimum score she needs to make on fifth test to achieve average of at least 80%.
Solution:
Let the minimum score she needs to make in fifth test be 'x'.
Total number of test = 5
Now we know that;
Average is equal to sum of all the scores in the test divided by number of test.
framing in equation form we get;
![\frac{87+92+76+89+x}{5}=80](https://tex.z-dn.net/?f=%5Cfrac%7B87%2B92%2B76%2B89%2Bx%7D%7B5%7D%3D80)
Multiplying both side by 5 we get;
![\frac{344+x}{5}\times 5=80\times 5\\\\344+x=400](https://tex.z-dn.net/?f=%5Cfrac%7B344%2Bx%7D%7B5%7D%5Ctimes%205%3D80%5Ctimes%205%5C%5C%5C%5C344%2Bx%3D400)
Subtracting both side by 344 we get;
![344+x-344=400-344\\\\x=56\%](https://tex.z-dn.net/?f=344%2Bx-344%3D400-344%5C%5C%5C%5Cx%3D56%5C%25)
Hence Sandra need to score at least <u>56%</u> in her fifth test so that her average is 80%.