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AnnyKZ [126]
3 years ago
7

Okay google how many pennies are in $25,000

Mathematics
1 answer:
GalinKa [24]3 years ago
4 0
There are 2,500,000 pennies in $25,000
You might be interested in
What are the solutions to 2x^2+7y=4 ? Select all that apply.
liberstina [14]

Answer:

x = -1/2 , 4

Step-by-step explanation:

1) If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

2x+1=0

x−4=0

2) Set the first factor equal to 0.

2x+1=0

3) Subtract 1 from both sides of the equation.

2x=−1

4) Divide each term by 2 and simplify.

2x/2= −1/2

5) Cancel the common factor of 2.

x= −1/2

6) Move the negative in front of the fraction.

x = −1/2

7) Set the next factor equal to 0 and solve.

x=4

8) The final solution is all the values that make

(2x+1)(x−4) = 0 true.

9) x = −1/2 , 4

7 0
3 years ago
I need B.<br><br> EMERGENCY THANK YOU VERY MUCH!!!
lisov135 [29]

Answer:

A. Mean = 9.7 (take all the numbers and divide by 10)

Mode = 12

Range = 5

B. Unsure. Would suggest either Mean - average and can tell when bus is running late or early or Mode - highest and lowest time combined

7 0
3 years ago
Chase is making calzones that require 2 3/4 pounds of beef. He only has 1,1/6 pound left. How much more does he need?
Dovator [93]
Here is the solution of the given problem above.
Firstly, to make the solution easier, let us convert the mixed fractions.
2 3/4 is 11/4 (Pounds of beef needed)
1 1/6 is 7/6 (Pounds of beef available)
So, to get how much more he needs, we will deduct 7/6 from 11/4 and we get 19/12. And this fraction is equivalent to option A. 1 7/12. Therefore, he needs 1 7/12 pounds more. hope this answer helps.

6 0
2 years ago
Read 2 more answers
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
2 years ago
A rectangle has a perimeter 60m and base21 m what is its area
Komok [63]

Since you're given the base for the perimeter of the rectangle, you can solve this question quickly.

First, you multiply 21 by 2, because there are two bases to a rectangle.

You'll get 42, and you will have to do 60-42 to find the perimeter for the remaining two sides.

60-42 will give you 18, so you can divide that by 2 because there are two other sides, giving you 9.

Now you have the side lengths 21 and 9. Area is base*height, so you can multiply 21*9 and get 189.

Therefore, the area of the rectangle is 189 meters

5 0
3 years ago
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