We know that 1 serving = 4 bananas
3 apples
6 pears
and 20 strawberrys
To get 30 we need to half 20 =10 20+10=30
That means we also needs to half the other ones
1/2 of 4 = 2 4+2=6
1/2 of 3 = 1.5 3+ 1.5=4.5
1/2 of 6 = 3 6+3=9
6 Bananas
4.5 Apples
9 Pears
30 strawberries diced.
Add the values, then you get the answer.
55+67 = ?
55
+ 67
------------
122
Final answer: $122 is how much Jordan has.
Answer:
Therefore, the four intervals are
(1) 6 - 6.59
(2) 7 - 7.59
(3) 8 - 8.59
(4) 9 - 9.59
The four frequencies are
(1) 4
(2) 3
(3) 1
(4) 6
Step-by-step explanation:
From the data, we have
Interval Frequency
1st 6 - 6.59 4
2nd 7 - 7.59 3
3rd 8 - 8.59 1
4th 9 - 9.59 6
Therefore, the four intervals are
(1) 6 - 6.59
(2) 7 - 7.59
(3) 8 - 8.59
(4) 9 - 9.59
The four frequencies are
(1) 4
(2) 3
(3) 1
(4) 6
Answer:
Step-by-step explanation:
0.3
^ You need to move the three before the decimal place and leave 4 empty digits (0) between
0.3
3.0
30.0
300.0
3000.0
30000.0
A total of 5 moves.
Each move is equivalent to x, the preceding value, times 10. x * 10
Repeated multiplication times the same factor equates to exponentiation, so for a total of 5 moves, it is notated as , giving you an expression of:
Answer:
The largest annual per capita consumption of bananas in the bottom 5% of consumption is 5.465 lb
Step-by-step explanation:
Given
μ = Mean = 10.4 lb
σ = Standard deviation = 3 lb
Using a confidence level of 90%,
We'll need to first determine the z value that correspond with bottom 5% of consumption of banana
α = 5%
α = 0.05
So,
zα = z(0.05)
z(0.05) = -1.645 ----- From z table
Let x represent the largest annual per capita consumption of bananas
The relationship between x and z is
x = μ + zσ
By substitution;
x = 10.4 + (-1.645) * 3
x = 10.4 - 4.935
x = 5.465
Hence, the largest annual per capita consumption of bananas in the bottom 5% of consumption is 5.465 lb