25% each or .25 her plus three friends =4 /4 =1 /
Answer:
Yes. The data provide enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
P-value=P(t>2.84)=0.0024
Step-by-step explanation:
Hypothesis test on the population mean.
The claim is that the mean weight of one-year-old boys is greater than 25 pounds.
Then, the null and alternative hypothesis are:

The significance level is α=0.05.
The sample size is n=354. The sample mean is 25.8 pounds and the sample standard deviation is 5.3 pounds. As the population standard deviation is estimated from the sample standard deviation, we will use a t-statistic.
The degrees of freedom are:

The t-statistic is:

For a right tailed test and 353 degrees of freedom, the P-value is:

As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.
There is enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
Answer: 105 couple tickets and 40 individual tickets
For this I will use the "Graphing Method"
First, we need to set up two equations.
-> Let x be the number of couple tickets.
-> Let y be the number of individual tickets.
-> It is "2x" because a couple means two.
$12x + $8y = $1,580
2x + y = 250
Next, we will graph this.
-> The point at which the two lines intersect is our solution.
-> See attached.
-> x = 105, so 105 couple tickets
-> y = 40, so 40 individual tickets
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
Answer:
a) The continuous rate of growth of this bacterium population is 45%
b) Initial population of culture at t = 0 is 950 bacteria
c) number of bacterial culture contain at t = 5 is 9013 bacteria
Step-by-step explanation:
The number of bacteria in a culture is given by
n(t) = 950
a) rate of growth is
Bacteria growth model N(t) = no
Where r is the growth rate
hence r = 0.45
= 45%
b) Initial population of culture at t = 0
n(0) = 950
= 950
= 950 bacteria
c) number of bacterial culture contain at t = 5
n(5) = 950
= 950
= 9013.35
= 9013 bacteria