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Klio2033 [76]
3 years ago
5

Can someone help this is division with remanders

Mathematics
1 answer:
Talja [164]3 years ago
3 0
Not really good at that sorry but one remaders are left over from the problem

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Tyre measured the average monthly temperature in degrees Fahrenheit for several months. If February is set ect 2 and June is oud
Katena32 [7]

Answer:

it will be d

Step-by-step explanation:

7 0
3 years ago
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Using the distance formula, find the distance between two points. (6, 0) (-5, 4)
Ipatiy [6.2K]

Answer:

11.7046999

Step-by-step explanation:

using the distance formula (d=√(x₂ -x₁)² + (y₂-y₁)²), we can plug in the two points to find the distance between them:

d=√(6-(-5)² + (0-4)²

d=√(11)²+ (-4)²

d=√121+16

d=√137

d approximately equal to (≈) 11.7046999 or 11.7 or 11

3 0
4 years ago
John painted 4 walls in 3 hours. Write an equation comparing walls painted to hours.
V125BC [204]
I think you should go with A).
3 0
3 years ago
Lucio quiere repartir su colección de estampillas entre la mayor cantidad de amigos pero de manera que todos reciban la misma ca
Daniel [21]

Answer:

Lucio podrá armar 5 grupos iguales de estampillas.

Step-by-step explanation:

El problema plantea un caso de distribución mediante la aplicación del divisor común mayor a todos los números involucrados. Para ello, primero debemos analizar los divisores de cada uno de estos números:

-30: divisible por 2, 3, 5, 6, 10, 15 y 30.

-75: divisible por 3, 5, 15, 25 y 75.

-160: divisible por 2, 4, 5, 8, 10, 16, 20, 32, 40, 80 y 160.

Así, podemos ver que el divisor común mayor a los tres números es 5, ya que 30, 75 y 160 dan como resultado un número entero tras ser divididos por 5.

Entonces, para poder repartir equitativamente sus estampillas entre la mayor cantidad de amigos posibles, deberá repartir 6 estampillas de animales, 15 estampillas de flores y 32 estampillas de ciudades a 5 amigos. De esta manera, Lucio podrá armar 5 grupos iguales de estampillas.

4 0
4 years ago
A survey was done that asked people to indicate whether they preferred to swim in a pool or in an ocean. The results are shown i
posledela
Given:
                                 Pool swimming        Ocean swimming    Total
Age 30 and younger             228                         54                 282
Over 30 years old                142                        185                 327
total                                    370                        239                609

Frequencies:
                                Pool swimming        Ocean swimming    Total
Age 30 and younger     228/609 = 0.37         54/609 = 0.09       282/609 = 0.46
Over 30 years old        142/609 = 0.23       185/609 = 0.30       327/609 = 0.54
<span>total                           370/609 = 0.61       239/609 = 0.39       609/609 = 1.00</span>

<span>The relative frequency (rounded to the nearest hundredth) of a person over 30 years old who prefers to swim in the ocean is 0.30</span>    
8 0
3 years ago
Read 2 more answers
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