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3 years ago

We have 12 pumpkin pies that need to be shared evenly between 16 people

Mathematics

1 answer:

Artyom0805 [142]3 years ago

7 0

You can divide 16 and 12, which will get you 1 and 1/4 a person left, but rounded will get 2

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(Ch. 6, p. A man bought a trampoline for his children to use. The trampoline has a diameter of 17 feet. If the man wishes to put

iVinArrow [24]

Diameter D of the trampoline = 17 feet

The circumference C, of the trampoline is calculated by the formula:

$C=\pi D$

Put D = 17 feet, π=3.14

$\begin{gathered} C=3.14\times17\text{feet} \\ C=53.38\text{feet} \end{gathered}$

The cirucmference o

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10 months ago

A box contains 4 black shirts, 8 blue shirts, 4 black pants, and 10 blue pants. Determine the probability of randomly selecting

Elenna [48]

So there is 26 shirts in total because 4+8+4+10=26 Then you need to take 26-4=22. You do this because there is 26 total clothing and only 4 of them aren't either a blue piece of clothing or pants that means you have 22 pieces of clothing that are blue or pants. So your odds of picking a blue piece of clothing or pants is 22/26.

Hope this Helps!!

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3 years ago

After you prove that two triangles are congruent, what would CPCTC be useful for?

Lapatulllka [165]

Is useful in provingvarious theorems about triangles and other polygons.

6 0

3 years ago

At the bottom is a completed addition problem, with all the digits replaced by letters. Every letter represents a single digit a

Rashid [163]

Answer:

T = 5 (H = 7, M = 1); T = 7 (H = 4, M = 2); T = 9 (H = 1, M = 3)

Step-by-step explanation:

We know MH+MH+MH=TM

MH is a two digit number, and TM is also a two digit number.

MH+MH+MH=3*(MH)<100

3M<=9, meaning that M could be 1, 2, or 3.

If M=1, MH+MH+MH=3(10+H)=30+3H=TM=10*T+1

30+3H=10T+1, so 3H must include 1 in one's digit.

The only possibility is that 3H = 21, then H = 7

MH+MH+MH=17+17+17=TM=51, T = 5

If M=2, MH+MH+MH=3(20+H)=60+3H=TM=10T+2

60+3H=10T+2, so 3H must include 2 in one's digit.

The only possibility is 3H = 12, then H = 4

MH+MH+MH=24+24+24=72, T = 7

If M=3, MH+MH+MH=3(30+H)=90+3H=TM=10T+3

90+3H=10T+3, so 3H must include 3 in one's digit.

The only possibility is 3H = 3, then H =1

MH+MH+MH=31+31+31=93, T=9

4 0

3 years ago

The letters in the word STATISTICS are printed on identical index cards with one letter per card. The ten cards are

Alex17521 [72]

Answer:

Outcome ---------- Probability

S -------------------------- 0.3

T -------------------------- 0.3

A -------------------------- 0.1

I --------------------------- 0.2

S -------------------------- 0.1

$P(V) = 2/10$

$P(F) = 2/10$

Events V and F are not mutually exclusive

Step-by-step explanation:

Given

$Text: STATISTICS$

Solving (a): Outcomes and Probabilities

The text is made of the following makeup.

$S = 3$

$T = 3$

$A = 1$

$I = 2$

$C = 1$

$Total = 10$

The probability of each outcome is calculated by dividing the occurrence of each alphabet by the total number of alphabets

So:

$S = 3/10 = 0.3$

$T = 3/10 = 0.3$

$A = 1/10 = 0.1$

$I = 2/10 = 0.2$

$C = 1/10 = 0.1$

In tabular form, we have:

Outcome ---------- Probability

S -------------------------- 0.3

T -------------------------- 0.3

A -------------------------- 0.1

I --------------------------- 0.2

S -------------------------- 0.1

Solving (b)

$V = Vowels$

$F = A - M$

Calculating P(V)

P(V) = Number of vowels divided by total

The vowels are A and I.

So, number of vowels is 2

$P(V) = 2/10$

Calculating P(F)

P(F) = Number of alphabets from A to M divided by total

Alphabets from A and M are A and I

So, number of alphabets in this category is 2

So:

$P(F) = 2/10$

Solving (c):

Events V and F are not mutually exclusive because both can occur at the same time.

i.e.

An event can be a vowel and still be between A and M

3 0

2 years ago