A chef bought 0.56 kilograms of almonds and 0.4 kilograms of pecans. How many kilograms of nuts did the chef buy in all?
1 answer:
You might be interested in
Answer:
They will win 48 games in a 120 game season.
Step-by-step explanation:
The team won 20 games out of 50; that is a 20:50 ratio, that is equal to x:120. x is the amount of games won in the 120 game season.
Set up the ratios like this:
=
cross multiply the numerators and denominators to get this:
20*120 = 50x
solve for x:
2,400 = 50x
x = 48
Answer:
C) 2,6,24,120,720
Step-by-step explanation:
Here the n-th term An =
A1 is the first term
A1 = (1 + 2)!/(1 + 2)
= 3!/3
A1 = 2
A2 is the second term
A2 = (2 +2)!/(2 +2)
= 4!/4
A2 = (1*2*3*4) /4
A2 = 6
A3 is the third term
A3 = (3 + 2)!/(3 +2)
A3 = 5!/5
A3 = 24
A4 is the fourth term
A4 = (4 + 2)!/(4 + 2)
A4 = 6!/6
A4 = 120
A5 is the fifth term
A5 = (5 + 2)!/(5 +2)
A5 = 7!/7
A5 = 720
Answer: C) 2,6,24,120,720
Thank you.
Answer:
P-value (t=2.58) = 0.0066.
Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean μ exceeds 2.4.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=45.
The sample mean is M=2.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the population mean μ exceeds 2.4.