Assuming we start with a full standard deck of 52 and then draw 4 spades along with 1 card from a different suit, that leaves us with 47 cards still in the deck. We are hoping to draw another spade from these remaining cards. Initially, there were 13 spades, but after drawing five cards, 9 spades remain in the deck. The likelihood of pulling one of these 9 spades from the 47 cards is
In other words, we want to get any 1 of the available 9 spades while avoiding any of the other 38 non-spades, and we're drawing just a single card from the 47 cards total.
Answer:
Option 2 50 ≤ s ≤ 100
Option 5 She can make a deposit of $50
Option 6 She can make a deposit of $75
Detailed explanation:
Let
s represent the amount of money Layla puts into her savings account.
We know that
25% = 25/100 = 0.25
50% = 50/100 = 0.50
Thus
-----> 
-----> 
The compound inequality is
Check each case
case 1) 25 ≤ s ≤ 50
This statement is false
Refer to the procedure
case 2) 50 ≤ s ≤ 100
This statement is true
Refer to the procedure
case 3) s ≤ 25 or s ≥ 50
This statement is false
Because s ≤ 100 and s ≥ 50
case 4) s ≤ 50 or s ≥ 100
This statement is false
Because s ≤ 100 and s ≥ 50
case 5) She can make a deposit of $50
This statement is true
As the value of s meets the compound inequality
case 6) She can make a deposit of $75
This statement is true
As the value of s meets the compound inequality
A) The cost to send a package that weighs 3.2 pounds is $4.13. Since this weight exceeds 3 pounds but remains below 4 pounds, we have to refer to the pricing that applies to 4-pound packages (see the attached document for pricing details).
b) To illustrate the Media Mail shipping costs based on the weight of the books, a line graph is appropriate. In this graph, the weight in pounds is represented on the x-axis and the shipping costs on the y-axis.
c) The graph depicting the Media Mail shipping costs as a function of book weight will be represented by the equation: f(x) = 2.69 + 0.48(x-1)
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
The answer is C. Explanation: The shirt originally priced at $35 has a discount of 15%. Therefore, the savings amount will be calculated accordingly.
