Answer:
round it to the nearest whole number.
Step-by-step explanation:
Answer:
x=27
Step-by-step explanation:
1. x-2=25 so add 2 to both sides x=25+2
2. 25+2= 27 so, the answer is x=27 Can you make me Brainliest
For this case we must solve each of the functions.
We have then:
f (x) = x2 - 9, and g (x) = x - 3
h (x) = (x2 - 9) / (x - 3)
h (x) = ((x-3) (x + 3)) / (x - 3)
h (x) = x + 3
f (x) = x2 - 4x + 3, and g (x) = x - 3
h (x) = (x2 - 4x + 3) / (x - 3)
h (x) = ((x-3) (x-1)) / (x - 3)
h (x) = x-1
f (x) = x2 + 4x - 5, and g (x) = x - 1
h (x) = (x2 + 4x - 5) / (x - 1)
h (x) = ((x + 5) (x-1)) / (x - 1)
h (x) = x + 5
f (x) = x2 - 16, and g (x) = x - 4
h (x) = (x2 - 16) / (x - 4)
h (x) = ((x-4) (x + 4)) / (x - 4)
h (x) = x + 4
Answer:
x = friends that paid discount price = 8
y = friends that paid regular price = 4
Step-by-step explanation:
Let
x = friends that paid discount price
y = friends that paid regular price
x + y = 12 (1)
6x + 8y = 80 (2)
From (1)
x = 12 - y
Substitute x = 12 - y into (2)
6x + 8y = 80 (2)
6(12 - y) + 8y = 80
72 - 6y + 8y = 80
- 6y + 8y = 80 - 72
2y = 8
y = 8/2
y = 4
Substitute y = 4 into (1)
x + y = 12 (1)
x + 4 = 12
x = 12 - 4
x = 8
x = friends that paid discount price = 8
y = friends that paid regular price = 4
Answer:
C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.
Step-by-step explanation:
Let μ be the the average attendance at games of the football team
The claim: the average attendance at games is over 523
Null and alternative hypotheses are:
: μ=523
: μ>523
The conclusion is failure to reject the null hypothesis.
This means that <em>test statistic</em> is lower than <em>critical value</em>. Therefore it is not significant, there is no significant evidence to accept the <em>alternative</em> hypothesis.
That is no significant evidence that the average attendance at games of the football team is greater than 523.
