Answer: b. 134
Step-by-step explanation:
Given : A minimum usual value of 135.8 and a maximum usual value of 155.9.
Let x denotes a usual value.
i.e. 135.8< x < 155.9
Therefore , the interval for the usual values is [135.8, 155.9] .
If interval for any usual value is [135.8, 155.9] , then any value should lie in this otherwise we call it unusual.
Let's check all options
a. 137 ,
since 135.8< 137 < 155.9
So , it is usual.
b. 134
since 134<135.8 (Minimum value)
So , it is unusual.
c. 146
since 135.8< 146 < 155.9
So , it is usual.
d. 155
since 135.8< 1155 < 155.9
So , it is usual.
Hence, the correct answer is b. 134 .
Look it up on the internet youll find the answer there i am only learining about adding integers right now
60.
let x be the number you're trying to solve.
(x/100)25 = 15
x/4 = 15
x = 15(4)
x = 60
The y intercept of this function is always (0, a).
This is because when we place 0 in for x (which is the only way it'll be on the y-axis, we get 'a' as a result. This is because of the rule that raising anything to the 0th power will result in the number 1 and multiplying anything by 1 gives us the same number. See the work below for the example.
F(x) = a*b^x
F(0) = a*b^0
F(0) = a*1
F(0) = a
And for an example with a random number, we'll use a = 5 and b = 3
F(x) = 5*3^x
F(0) = 5*3^0
F(0) = 5*1
F(0) = 5
No matter what a and b equal, the intercept will be the a value.
Answer:
<em>We can't find a unique price for an apple and an orange.</em>
Step-by-step explanation:
Suppose, the price of an apple is 
and the price of an orange is 
They need $10 for 4 apples and 4 oranges. So, the first equation will be.......
They also need $15 for 6 apples and 6 oranges. So, the second equation will be........
Dividing equation (1) by 2 on both sides : 
Dividing equation (2) by 3 on both sides : 
So, we can see that both equation (1) and (2) are actually same. That means, we will not get any unique solution for and here. Both and have <u>"infinitely many solutions"</u>.
Thus, we can't find a unique price for an apple and an orange.
