All categories

3 years ago

The grocery sells beans in bulk. The grocer's sign above the beans says, "5 pounds for $4."

Mathematics

1 answer:

solong [7]3 years ago

4 0

Answer:

A). Alberto is the only right one because .80 * 5 = 4$

B). Claude will need 1.60$ because .80 * 2 = 1.60$

C). Dora can buy 12.5 pounds because 10$ / .80 is 12.5

D). Using Alberto's because he was the one who was correct

Step-by-step explanation:

please leave brainiest 5 star and thx i really appreciate it

You might be interested in

Last week, Arnold purchased 3 apples at the store for $3.90. This week he purchased 5 apples for $6.50. Help Arnold analyze the

Minchanka [31]

<span>Last week he bought three apples for $3.90. The cost of an apple = 3.9/3 = 1.3 This week he bought 2 more apples which totals 5 for $6.50 The cost of each apple = 6.5/5 = 1.3. The cost of an apple in the two weeks is the same. Arnold bought more apples in the following week but there was a corresponding raise in price. When he bought fewer apples last week he never had to pay as much as he did the year later</span>

6 0

3 years ago

Help pleaseee hurry if u can

Alborosie

Answer:

sorry your questions image is blurr

3 0

3 years ago

-7+ 5Y = -37 simplify your answer as much as possible

irinina [24]

Answer:

y = -6

Step-by-step explanation:

To simplify this equation as much as possible, we have to get the variable y by itself.

To do this, we would do the inverse operation of subtraction and add 7 to the left and right side of the equation.

-7 + 7 + 5y = -37 + 7

The 7s cancel out on the left side because -7+7 = 0 and so we are left with,

5y = -30

Then we would do the inverse operation of multiplication and divide by 5 on the left and right side of the equation.

5y/5 = -30/5

y = -6

8 0

3 years ago

Thank you so much to anyone who answers this question!

Anna71 [15]

C. The 40th customer

4 0

3 years ago

How many permutations of the 26 letters of the English alphabet do not contain any of the strings fish, rat, or bird

NARA [144]

The number of permutations of the 26 letters of the English alphabet that do not contain any of the strings fish, rat, or bird is 402619359782336797900800000

Let

$\mathcal{E}=\{\text{All lowercase letters of the English Alphabet}\}\\\\B=\overline{\{b,i,r,d\}} \cup \{bird\}\\\\F=\overline{\{f,i,s,h\}} \cup \{fish\}\\\\R=\overline{\{r,a,t\}} \cup \{rat\}\\\\FR=\overline{\{f,i,s,h,r,a,t\}} \cup \{fish,rat\}$

Then

$Perm(\mathcal{E})=\{\text{All orderings of all the elements of } \mathcal{E}\}\\\\Perm(B)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing bird}\}\\\\Perm(F)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing fish}\}\\\\Perm(R)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing rat}\}\\\\Perm(FR)=\{\text{All orderings of all the elements of } \mathcal{E} \text{ containing both fish and rat}\}\\$

Note that since

$F \cap R=\varnothing$ , $Perm(F)\cap Perm(R)\ne \varnothing$

But since

$B \cap R \ne \varnothing$ , $Perm(B)\cap Perm(R)= \varnothing$

and

$B \cap F \ne \varnothing$ , $Perm(B)\cap Perm(F)= \varnothing$

Since

$|\mathcal{E} |=26 \text{, then, } |Perm(\mathcal{E})|=26! \\\\|B|=26-4+1=23 \text{, then, } |Perm(B)|=23!\\\\|F|=26-4+1=23 \text{, then, } |Perm(F)|=23!\\\\|R|=26-3+1=24 \text{, then, } |Perm(R)|=24!\\\\|FR|=26-7+2=21 \text{, then, } |Perm(FR)|=21!\\$

where $|Perm(X)|=\text{number of possible permutations of the elements of X taking all at once}$

and

$|Perm(F) \cup Perm(R)| = |Perm(F)| + |Perm(R)| - |Perm(FR)|\\= 23!+24!- 21! \text{ possibilities}$

What we are looking for is the number of permutations of the 26 letters of the alphabet that do not contain the strings fish, rat or bird, or

$|Perm(\mathcal{E})|-|Perm(B)|-|Perm(F)\cup Perm(R)|\\= 26!-23!-(23!+24!- 21!)\\= 402619359782336797900800000 \text{ possibilities}$

This link contains another solved problem on permutations:

brainly.com/question/7951365

6 0

3 years ago