Help in what ? I cant see the photo
Im assuming you mean 20pence? if so its 5 twenty pence in a pound so times 5 by 9 gives you 45
so there are 45 twenty pence in £9
Answer:
95% provides more information
Step-by-step explanation:
The confidence interval is obtained by using the relation :
Xbar ± Zcritical * σ/√n
(Xbar - (Zcritical * σ/√n)) = 5.22 - - - (1)
(Xbar + (Zcritical * σ/√n)) = 5.98 - - (2)
Adding (1) and (2)
2xbar = 5.22 + 5.98
2xbar = 11.2
xbar = 11.2 / 2 = 5.6
Margin of Error :
Xbar - lower C.I = Zcritical * σ/√n
Zcritical at 90% = 1.645
5.6 - 5.22 = 1.645 * (σ/√n)
0.38 = 1.645 * (σ/√n)
(σ/√n) = 0.38 / 1.645 = 0.231
Therefore, using the se parameters to construct at 95%
Zcritical at 95% = 1.96
Margin of Error = Zcritical * σ/√n
Margin of Error = 1.96 * 0.231 = 0.45276
C.I = xbar ± margin of error
C. I = 5.6 ± 0.45276
C.I = (5.6 - 0.45276) ; (5.6 + 0.45276)
C. I = (5.147 ; 6.053)
Hence, 95% confidence interval provides more information as it is wider.
Answer:
after 75 minutes
Step-by-step explanation:
The least common multiple (LCM) of 15 and 25 is 75. It can be found a couple of ways:
1. List the factors of each number and find the product of the unique ones:
15 = 3·5
25 = 5²
The LCM is 3·5² = 75.
__
2. Find the greatest common divisor (GCD) and divide the product of the numbers by that value. From the above list of factors, we see that 5 is the GCD of 15 and 25. Then the LCM is ...
15·25/5 = 75
__
Or, you can simply list multiples of each number and see what the smallest number is that is in both lists:
15, 30, 45, 60, <em>75</em>, 90
25, 50, <em>75</em>, 100
__
The two buses will appear together again after 75 minutes.
Answer: The number is 26.
Step-by-step explanation:
We know that:
The nth term of a sequence is 3n²-1
The nth term of a different sequence is 30–n²
We want to find a number that belongs to both sequences (it is not necessarily for the same value of n) then we can use n in one term (first one), and m in the other (second one), such that n and m must be integer numbers.
we get:
3n²- 1 = 30–m²
Notice that as n increases, the terms of the first sequence also increase.
And as n increases, the terms of the second sequence decrease.
One way to solve this, is to give different values to m (m = 1, m = 2, etc) and see if we can find an integer value for n.
if m = 1, then:
3n²- 1 = 30–1²
3n²- 1 = 29
3n² = 30
n² = 30/3 = 10
n² = 10
There is no integer n such that n² = 10
now let's try with m = 2, then:
3n²- 1 = 30–2² = 30 - 4
3n²- 1 = 26
3n² = 26 + 1 = 27
n² = 27/3 = 9
n² = 9
n = √9 = 3
So here we have m = 2, and n = 3, both integers as we wanted, so we just found the term that belongs to both sequences.
the number is:
3*(3)² - 1 = 26
30 - 2² = 26
The number that belongs to both sequences is 26.
