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Amanda [17]
3 years ago
11

A recipe for a loaf of bread calls for 2/3 of a cup of flour. If Milo used 12 cups of flour, how many loafs of bread did he prep

are?
Mathematics
2 answers:
Aloiza [94]3 years ago
7 0
The answer is:

1/2 cups of flour can make 1 loaf.

1 cup would make =   1 /(1/2) = 2 loaves.

12 cups of flour would then make =  12*2 = 24

= 24 loaves.

[Please Mark as Brainliest]

KatRina [158]3 years ago
3 0

ratio and proportion

1 loaf is to 2/3 cup

? loaves is to 12 cups

1:2/3=?:12

so

\frac{1}{\frac{2}{3}}=\frac{?}{12}

cross multiply or basically multipily both sides by (\frac{2}{3})(12)

12=\frac{2}{3}?

multiply both sides by \frac{3}{2}

\frac{36}{2}=?

18=?

18 loaves of bread was prepared

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What 2 numbers can you multiply to get 31? can 31 get into smaller groups?
zhannawk [14.2K]
Base on the question and on my further research and observation, I would say that the answer would be 1*31. I hope you are satisfied with my answer and feel free to ask for more if you have clarification and further questions about the said problem. 
7 0
3 years ago
List the probability value for each possibility in the binomial experiment calculated at the beginning of this lab, which was ca
Yanka [14]

Answer:

a. P(X = 0)= 0.001

b. P(X = 1)= 0.001

c. P(X=2)= 0.044

d. P(X=3)= 0.117

e. P(X=4)= 0.205

f. P(X=5)= 0.246

g. P(X=6)= 0.205

h. P(X=7)= 0.117

i. P(X=8)= 0.044

j. P(X=9)= 0.001

k. P(X=10)= 0.001

Step-by-step explanation:

Hello!

You have the variable X with binomial distribution, the probability of success is 0.5 and the sample size is n= 10 (I suppose)

If the probability of success p=0.5 then the probability of failure is q= 1 - p= 1 - 0.5 ⇒ q= 0.5

You are asked to calculate the probabilities for each observed value of the variable. In this case is a discrete variable with definition between 0 and 10.

You have two ways of solving this excersice

1) Using the formula

P(X)= \frac{n!}{(n-X)!X!} * (p)^X * (q)^{n-X}

2) Using a table of cummulative probabilities of the binomial distribution.

a. P(X = 0)

Formula:

P(X=0)= \frac{10!}{(10-0)!0!} * (0.5)^0 * (0.5)^{10-0}

P(X = 0) = 0.00097 ≅ 0.001

Using the table:

P(X = 0) = P(X ≤ 0) = 0.0010

b. P(X = 1)

Formula

P(X=1)= \frac{10!}{(10-1)!1!} * (0.5)^1 * (0.5)^{10-1}

P(X = 1) = 0.0097 ≅ 0.001

Using table:

P(X = 1) = P(X ≤ 1) - P(X ≤ 0) = 0.0107-0.0010= 0.0097 ≅ 0.001

c. P(X=2)

Formula

P(X=2)= \frac{10!}{(10-2)!2!} * (0.5)^2 * (0.5)^{10-2}

P(X = 2) = 0.0439 ≅ 0.044

Using table:

P(X = 2) = P(X ≤ 2) - P(X ≤ 1) = 0.0547 - 0.0107= 0.044

d. P(X = 3)

Formula

P(X = 3)= \frac{10!}{(10-3)!3!} * (0.5)^3 * (0.5)^{10-3}

P(X = 3)= 0.11718 ≅ 0.1172

Using table:

P(X = 3) = P(X ≤ 3) - P(X ≤ 2) = 0.1719 - 0.0547= 0.1172

e. P(X = 4)

Formula

P(X = 4)= \frac{10!}{(10-4)!4!} * (0.5)^4 * (0.5)^{10-4}

P(X = 4)= 0.2051

Using table:

P(X = 4) = P(X ≤ 4) - P(X ≤ 3) = 0.3770 - 0.1719= 0.2051

f. P(X = 5)

Formula

P(X = 5)= \frac{10!}{(10-5)!5!} * (0.5)^5 * (0.5)^{10-5}

P(X = 5)= 0.2461 ≅ 0.246

Using table:

P(X = 5) = P(X ≤ 5) - P(X ≤ 4) = 0.6230 - 0.3770= 0.246

g. P(X = 6)

Formula

P(X = 6)= \frac{10!}{(10-6)!6!} * (0.5)^6 * (0.5)^{10-6}

P(X = 6)= 0.2051

Using table:

P(X = 6) = P(X ≤ 6) - P(X ≤ 5) = 0.8281 - 0.6230 = 0.2051

h. P(X = 7)

Formula

P(X = 7)= \frac{10!}{(10-7)!7!} * (0.5)^7 * (0.5)^{10-7}

P(X = 7)= 0.11718 ≅ 0.1172

Using table:

P(X = 7) = P(X ≤ 7) - P(X ≤ 6) = 0.9453 - 0.8281= 0.1172

i. P(X = 8)

Formula

P(X = 8)= \frac{10!}{(10-8)!8!} * (0.5)^8 * (0.5)^{10-8}

P(X = 8)= 0.0437 ≅ 0.044

Using table:

P(X = 8) = P(X ≤ 8) - P(X ≤ 7) = 0.9893 - 0.9453= 0.044

j. P(X = 9)

Formula

P(X = 9)= \frac{10!}{(10-9)!9!} * (0.5)^9 * (0.5)^{10-9}

P(X = 9)=0.0097 ≅ 0.001

Using table:

P(X = 9) = P(X ≤ 9) - P(X ≤ 8) = 0.999 - 0.9893= 0.001

k. P(X = 10)

Formula

P(X = 10)= \frac{10!}{(10-10)!10!} * (0.5)^{10} * (0.5)^{10-10}

P(X = 10)= 0.00097 ≅ 0.001

Using table:

P(X = 10) = P(X ≤ 10) - P(X ≤ 9) = 1 - 0.9990= 0.001

Note: since 10 is the max number this variable can take, the cummulated probability until it is 1.

I hope it helps!

4 0
3 years ago
Megan is buying a new stove for 499.99. Sales tax is 7.5%. How much will Meagan have to pay in total for the stove? Round your a
bija089 [108]
I don't understand this question
4 0
3 years ago
Read 2 more answers
I NNEEEDDDDDD HELPPPPP PLEASEEEE
NikAS [45]

Answer:

The small balloon bouquet uses  7 balloons and the large one uses

18 balloons.

Step-by-step explanation:

Let's say that small balloon bouquets are S and large balloon bouquets are L. For the graduation party the employee assembled 6 small bouquets and 6 large bouquets, the total number of balloon used is 150. To put the sentence into an equation will be:

6S + 6L= 150

S+L= 25   ----> 1st equation

For Father's Day, the employee uses 6 small bouquet and 1 large bouquet, the total number of balloons used is 60. The equation will be:

6S + 1L= 60

1L= 60- 6S   ----> 2nd equation

We can solve the number of small balloon bouquet by substitute the 2nd equation into 1st. The calculation will be:

S+L = 25

S+ (60-6S)= 25

-5S= 25-60

-5S= -35

S= -35/-5

S=7

Then we can find L by substitute S value to 1st or 2nd equation.

S+L=25

7+L=25

L=18

Hope this helps ;)

3 0
3 years ago
Read 2 more answers
Help please thanks quick
andrew11 [14]

Answer:

The last option has another result than the others.

Step-by-step explanation:

The first option answer is 7/2

The second option answer is 7/2

The third option answer is 7/2

The last option answer is 2/7

Hope it will help :)

6 0
3 years ago
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