Answer:
<h2>
<em><u>$23.40 </u></em></h2>
Step-by-step explanation:
<em><u>Given,</u></em>
Cost price of 8 cakes = $62.40
<em><u>So,</u></em>
Cost price of 1 cake = $ 
<em><u>Therefore,</u></em>
Cost price of 3 cakes
= $( × 3)
- <em>[On dividing 62.40 by 8]</em>
= $(7.80 × 3)
- <em>[On multiplying]</em>
= $23.40
<em><u>Hence,</u></em>
<em><u>The required cost of 3 cakes is $23.40 (Ans)</u></em>
<span>Proportional Relationship should be the correct answer.</span>
Answer:
why they're so helpful
Step-by-step explanation:
Answer:
a. 0.0368
b. 0.99992131
c. 0.2039
d. 0.0048
e. 0.6533
Step-by-step explanation:
Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35
Since this is a binomial probability, we use a binomial probability.
a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368
b. Probability of 2 or more heads P(x ≥ 2) is
P(x ≥ 2) = 1 - P(x ≤ 1)
Now P(x ≤ 1) = P(0) + P(1)
= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹
= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹
= 0.000003379 + 0.00007531
= 0.0007869
P(x ≥ 2) = 1 - P(x ≤ 1)
= 1 - 0.00007869
= 0.99992131
c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039
d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048
e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴
= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴
= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367
= 0.6533
Decimals that contain a zero in the tenth position can be thought of as a fraction over 100 as the tenth isn't occupied and the hundredth correlates to the 100 as a denominator. The same goes for a number with the tenth position occupied. These decimals would be put over 10 as they are occupying that position in the decimal.
