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Tamiku
7 days ago
9

After a (not very successful) trick or treating round, Candice has 15 Tootsie rolls and 9 Twizzlers in her pillow case. Her moth

er asks her to share some of the loot with her three younger brothers.
(A) How many different ways can she do this?

(B) How many different ways can she do this after her Mother asks her to give at least one of each type of candies to each of her brothers?
Mathematics
1 answer:
tester [11.9K]7 days ago
6 0

Solution:

There are 4 ways.

Detailed explanation:

Candice has a total of 15 + 9 = 24 candies. Since she has three younger brothers, and 24 can be divided by 3 (24/3 = 8). Both 15 and 9 can also be divided by 3 (15/3 = 5 and 9/3 = 3).

- She can distribute 5 tootsie rolls to each brother.

- She can provide 3 twizzlers to each brother.

- She can give each brother 5 tootsie rolls and 3 twizzlers (if she decides to share all her candies).

- She can give them one of each type of candy, leaving her with 12 tootsie rolls and 6 twizzlers (this would be the best option if she wants to keep some for herself).

I see four methods to accomplish this, and two methods remain after her mother instructs her to share at least one of each candy type with all three brothers.

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Shane and Abha earned a team badge that required their team to collect no less than 20002000 cans for recycling. Abha collected
tester [11953]

Shane and Abha received a team badge for gathering at least 2000 cans for recycling.

This indicates that their collection must total a minimum of 2000 cans.

Abha managed to collect 178 more cans than Shane.

Let’s denote the number of cans Shane collected as S

So, Abha collected = S + 178

The inequality representing the number of cans collected by Shane can be expressed as:

S+S+178\geq 2000

= 2S+178\geq 2000

2S\geq 2000-178

2S\geq 1822

S\geq 911

3 0
16 days ago
A)
Svet_ta [12361]
The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.
3 0
19 days ago
A child designs a flag in the shape of a rhombus, as shown in the diagram below. Which expression can be used to determine the s
Zina [12040]
<span>Which formula can be applied to find the side length of the rhombus?

 The correct answer is the first choice: 10/Cos(30°)

 Explanation:

 1. The figure shows a right triangle, where the hypotenuse is denoted by "x," and this is the length you are solving for. Therefore, you have:

 Cos(</span>α)=Adjacent side/Hypotenuse
<span>
 </span>α=30°
<span> Adjacent side=(20 in)/2=10 in
 Hypotenuse=x

 2. Inputting these numbers into the equation yields:
</span>
 Cos(α)=Adjacent side/Hypotenuse
<span> Cos(30°)=10/x

 3. Hence, by isolating the hypotenuse "x," you arrive at the expression to find the side length of the rhombus, as shown below:

 x=10/Cos(30°)
</span>

7 0
17 days ago
Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought
Svet_ta [12361]

Solution:

In Mr. Skinner's class, the count of students bringing lunch from home is 12 out of 20.

Fraction of students who brought lunch from home in Mr. Skinner's class=\frac{12}{20}=\frac{3}{5}

For Ms. Cho's class, the number who brought lunch from home is 14 out of 21.

Fraction of students who brought lunch from home in Ms. Cho's class=\frac{14}{21}=\frac{2}{3}

Siloni is utilizing two spinners with 15 equal sections to randomly select students from the classes and predict whether they brought lunch or will purchase it from the cafeteria.

Number of Equal sections in each Spinner=15

To visualize the students from Mr. Skinner's class who brought lunch using a Spinner with 15 equal sections =\frac{9}{15}

For Ms. Cho's class, using a Spinner with 15 equal sections =\frac{10}{15}

Mr. Skinner's Class +1 = Ms. Cho's Class

This means that the spinner for Ms. Cho's class will include one additional section representing students who brought lunch.

Option A signifies that one additional section on Mr. Skinner's spinner represents students who brought lunch, reflecting Ms. Cho's class.

9 0
1 month ago
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Simplify" (0.324/4.76+ 6.3)^-3​
Zina [12040]
I believe the answer is six.:)
5 0
11 days ago
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