The greatest common factor of the numbers is that which is the fused factors of all the numbers without the repetition of the common factors present in the given.
25·5·11 = 5·5·5·11
23·52·7 =23·2·2·13·7
Since, there are no repeated factors, the answer would be,
5·5·5·11·23·2·2·13·7
The answer unfortunately is not among the choices.
Answer:
steve=65 mph
kevin=13 mph
Step-by-step explanation:
steve=s
kevin=k
s=5
k=1
1=5=6
390 divided by 6 equals s 65
divide further by 5 which equals k 13
Answer:
x=-27
Step-by-step explanation:
-0.57x+0.27x = 8.1
-0.3x = 8.1
x = -8.1/0.3
x = -27
Answer:
Step-by-step explanation:
Hello!
Your study variable is X: "number of ColorSmart-5000 that didn't need repairs after 5 years of use, in a sample of 390"
X~Bi (n;ρ)
ρ: population proportion of ColorSmart-5000 that didn't need repairs after 5 years of use. ρ = 0.95
n= 390
x= 303
sample proportion ^ρ: x/n = 303/390 = 0.776 ≅ 0.78
Applying the Central Limit Theorem you approximate the distribution of the sample proportion to normal to obtain the statistic to use.
You are asked to estimate the population proportion of televisions that didn't require repairs with a confidence interval, the formula is:
^ρ±
* √[(^ρ(1-^ρ))/n]
= 
= 2.58
0.78±2.58* √[(0.78(1-0.78))/390]
0.0541
[0.726;0.834]
With a confidence level of 99% you'd expect that the interval [0.726;0.834] contains the true value of the proportion of ColorSmart-5000 that didn't need repairs after 5 years of use.
I hope it helps!
Answer:
108 student tickets, and 176 adult tickets were sold
Step-by-step explanation:
Adult ticket $8 Call the number of adult tickets sold "a"
Student ticket $5 Call the number of student tickets sold "s"
Since we are talking about TWO consecutive days of sold out seats, the total number of seats sold were 2* 142 = 284
Then we create two different equations with the information given:
a + s = 284
8 * a + 5 * s = 1948
we can solve for s in the first equation as follows: s = 284 - a
and use it in the second equation
8 a + 5 (284 - a) = 1948
8 a + 1420 - 5 a = 1948
combining
3 a = 528
a = 528/3
a = 176
we find the number of student tickets using this answer in the substitution equation we used:
s - 284 - 176 = 108
Therefore 108 student tickets, and 176 adult tickets were sold.
