Sounds like answer d to me
The answer is B (There is evidence that a Type BBB orange typically weighs more than a Type AAA orange.) this is right on edulastic
Answer:
Therefore,
Correct options are
A) 2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100
D) 2 × 100 + 6 × 1/100
Step-by-step explanation:
Given:
Number is
200.06
For option A)
2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100
= 200 + 0 + 0 + 0.06
= 200.06
Which is CORRECT.
For option B)
2 × 100 + 0 × 1 + 0 × 1/100 + 6 × 1/1,000
= 200 + 0 + 0 + 0.006
=200.006
Which is INCORRECT.
For option C)
2 × 100 + 6 × 1/10
= 200 + 0.6
= 200.6
Which is INCORRECT.
For option D)
2 × 100 + 6 × 1/100
= 200 + 0.06
= 200.06
Which is CORRECT.
Therefore,
Correct options are
A) 2 × 100 + 0 × 1 + 0 × 1/10 + 6 × 1/100
D) 2 × 100 + 6 × 1/100
Let's call the 13¢ stamps a and the 18¢ stamps b:
a+b = 42 and therefore a= 42-b (formula 1)
0.13a+0.18b= 6.66 In this formula, substitute the value of a according to formula 1:
0.13(42-b)+0.18b= 6.66 Multiply on the left to get rid of the parenthesis:
5.46-0.13b+0.18b= 6.66 Subtract 5.46 from both sides:
-0.13b+0.18b= 1.20 Add on the left:
0.05b= 1.20 Divide both sides by 0.05
b= 24 You have 24 18¢ stamps and:
42-24= 18 13¢ stamps
Check: (24 x 0.18) + (18 x 0.13)= 6.66 Correct.
(88 + 92 + 96 + x) / 4 = 90
(276 + x) / 4 = 90
276 + x = 90 * 4
276 + x = 360
x = 360 - 276
x = 84 <=== he would need an 84
