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Doss [256]
2 years ago
6

A display case of wallets are marked 32 for $5. If Tim has $80, how many wallets can Tim get? (Assume no tax or other fees.)

Mathematics
2 answers:
Levart [38]2 years ago
7 0
1 wallet= 5/32= 0.15625
Wallets for $80= 0.15625*80=12 wallets
UNO [17]2 years ago
4 0

Answer:

6.5 times 80 equals 512

Step-by-step explanation:

all you got to do is divide 32 by 5 then the answer times 80

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11/20 as a simplified fraction
NikAS [45]
Fraction already reduced simplified to lowest terms gcf ( 11/20 ) =1 numerator and denominator are coprime number they no common prime factors.
5 0
3 years ago
Mindy bought 5/8 pound of almonds and a 3/4 pound of walnuts.Which pair of fraction cannot be used to find how many pounds of nu
taurus [48]

Answer: 11 over 8 or 1 and 3 over 8

Step-by-step explanation:

3/4 is made into 6/8. Add 6/8 to 5/8 and you get 11/8 which is equal to 1 and 3/8.

6 0
4 years ago
How to put a fraction on lowest terms?
djverab [1.8K]

Answer: You simplify it.

Step-by-step explanation: You have to take the numerator and the denominator, find the greatest common factor, and divide it until it is at its simplest form.

Hope that helps

7 0
4 years ago
A sample of weights of defensive tackles on Big 10 football teams are: Johnson, 202 pounds; Patrick, 215 pounds; Junior, 207 pou
ExtremeBDS [4]

Answer:

the sample standard deviation is 4.926 pounds

Step-by-step explanation:

the sample standard deviation s is defined as

s =√[ 1/(n-1) ∑ ( x- m)²]  for x=1 to x=n

where

n= sample size = 10

x= weights of defensive tackles

m= sample mean

m= 1/n ∑ x*  for x=1 to x=n

then

m= 1/10 (202 + 215... +208) = 209.66 pounds

thus

s =√[ 1/(10-1) ∑ ( x- m)²] = √[ 1/(10-1) [ ( 202 - 209.66 )² + ( 215 - 209.66 )² +  ( 28 - 209.66 )² ] ] = 4.926 pounds

then the sample standard deviation is 4.926 pounds

4 0
3 years ago
For how many positive values of n are both n3 and 3n four-digit integers?
AVprozaik [17]

Answer:

No positive value of n

Step-by-step explanation:

we have to find out for how many positive values of n are both

n^3 , 3^n our-digit integers

Let us consider first cube

we get 4digit lowest number is 1000 and it has cube root as 10.

Thus 10 is the least integer which satisfies four digits for cube.

The highest integer is 9999 and it has cube root as 21.54

or 21 the highest integer.

Considering 3^n we get,

3^10 is having 5 digits and also 3^21

Thus there is no positive value of n which satisfy that both n cube and 3 power n are four digits.

5 0
3 years ago
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