Part 1: 4 quarts in a gallon
4 * 5 = 20 quarts
Part 2: 20 * 2 = 40
They will make $40.00
Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
The answer is false. This is because<u> 8 does not equal the number 28</u>.
Explination:
|8|
=
28
8≠28
Answer:(c)
Step-by-step explanation:
Given : In a hypothesis testing the null hypothesis has been rejected when the alternative hypothesis has been true.
Find : to find that the given decision is create type I error , type II error or right decision has been taken
Step by step :
In hypothesis testing there are 2 types of error.
(1)Type I error :- if the null hypothesis is rejected when it is true, is known has Type I error .
(2)Type II error:- if the null hypothesis is accepted when alternative hypothesis is true, is known as type II error.
Since here null hypothesis has been rejected when alternative hypothesis has been true,i.e., the correct decision has been made
Hence option (c) is correct.
Answer:
For the code we have 3 selections.
The first selection is a digit that must be odd, so the options are {1, 3, 5, 7 ,9}
So we have 5 options.
The second selection is a letter from the set of all the letters (27) minus the set of the vowels (5)
So here we have 27 - 5 = 22 options
The third selection is also a letter from the previous set, but because each letter can be used only one time, and in the previous selection we already selected one of the letters, in this selection we have a letter less than in the previous selection.
Here we have 22 - 1 = 21 options.
The total number of combinations (of possible codes) is equal to the product of the number of options for each selection:
C = 5*22*21 = 2310.
There are 2310 different possible codes
