Answer:
Yes
Step-by-step explanation:
<u>Answer:</u>
333 people of ward 5 are going to be voting for Spike Jones.
<u>Solution:</u>
We have been given that two-thirds of all voters in Ward 5 plan on choosing Spike Jones for commissioner.
There are 500 voters in Ward 5.
Since 2/3 of all voters in Ward 5 are voting for Spike Jones the remaining 1/3 will not be voting for him.
To find out how many people in ward 5 are exactly voting for Spike Jones. We need to calculate how much is two thirds of 500 is.
This is done as follows:
Since people cannot be denoted in decimal points we have to round it off to a whole number. That’s is 333.
Therefore 333 people of ward 5 are going to be voting for Spike Jones.
Find the soluiton
2x+2y=16
3x-y=4
x+y=8
<u>3x-y=4+</u>
4x=12
x=3
3(3)-y=4
9-y=4
y=5
(3,5)
test them to see if get true statment
obviously the first equatons of 1,2,4 work
1 doesn't work
2, works
4. doesn't work
2 is the same as the first except both euations are just doubled
answer is 2
Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.
Number of sample, n = 41
Mean, u = 43.1 hours
Standard deviation, s = 5.91 hours
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
43.1 ± 1.645 × 5.91/√41
= 43.1 ± 1.645 × 0.923
= 43.1 ± 1.52
The lower end of the confidence interval is 43.1 - 1.52 =41.58
The upper end of the confidence interval is 43.1 + 1.52 =44.62
Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
The answer is B, if you earn 200 points, then you receive 3 stars
