Ask question

Ask question

All categories

4 years ago

13

luis is making two batches of muffins for a school picnic. one batch of muffins uses 1/3 cup of oats and 1/2 cup of flour. how m

uch oats and flour does luis need for two batches? complete the explanation of how fraction strips are used to solve the problem

2 answers:

Llana [10]4 years ago

5 0

1 2/12 is the answer because you multiply the fraction by the other detonator

PolarNik [594]4 years ago

4 0

1/3+1/3=2/3 1/2+1/2=1 I'm not sure exactly what you're looking for. Let me know if this isn't what you're looking for.

You might be interested in

What would you do if a henchmen no scoped you 100000000 meters away in fortnite

nirvana33 [79]

Answer:

Smack my brother, because it was probably him.

5 0

3 years ago

My duvet cover is red . Statement or not

leva [86]

Answer:

It's definitely a statement

4 0

3 years ago

Read 2 more answers

A cone with volume 5000 m³ is dilated by a scale factor of 15. What is the volume of the resulting cone? Enter your answer in th

Rasek [7]

Answer:

$40\ m^{3}$

Step-by-step explanation:

we know that

If two figures are similar then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z-----> scale factor

x-----> the volume of the dilated cone

y-----> the volume of the original cone

$z^{3}=\frac{x}{y}$

In this problem we have

$z=1/5$

$y=5,000\ m^{3}$

substitute and solve for x

$(1/5)^{3}=\frac{x}{5,000}$

$(1/125)=\frac{x}{5,000}$

$x=5,000/125=40\ m^{3}$

5 0

3 years ago

Read 2 more answers

What is the missing rational expression in the following multiplication sentence (b³(a=7)/a^2)(?)=b(a-7)/14a

Nitella [24]

Answer:

C a/14b2

Step-by-step explanation:

8 0

2 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago