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3 years ago

Manuel created a factor tree and wrote the prime factorization of 60 shown below.

Mathematics

2 answers:

Galina-37 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

Manuel does the prime factorization as:

4 15

3 5

He placed the arrow under wrong number as he placed the arrow under 4 and 4 is not the prime number and can be factorized as 2, 2.

Also, the arrow should be placed under 15 as 3 and 5 are the factors of 15. Thus, the factor tree should in the form:

4 15

3 5

nata0808 [166]3 years ago

3 0

"Manuel" here put the arrow in the wrong place. Should be under 15, not 4. PLUS, 4 is not prime, cuz 2 goes into it...

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A loaf of bread costs $0.90. Each loaf has 15 slices. How much does each slice of bread cost?

scoray [572]

Answer:

Each slice of bread costs $0.06.

Step-by-step explanation:

To find how much each slice of bread costs, you need to divide the cost (0.90) by how many slices there are (15). When you divide the two, you get 0.06. To check your answer you can do 0.06x15 and that will equal $0.90.

8 0

3 years ago

Read 2 more answers

Using the following information = Universal set = {2,3,4,5,6,7,8,9,12,13,14,16,20,22,56). Subset A = {9,12,13,20,22,56); Subset

Cloud [144]

$\dfrac{2}{5},\dfrac{13}{15},\dfrac{3}{5},\dfrac{1}{5},\dfrac{1}{5}$

step-by-step explanation:

The intersection of two sets A and B, denoted by A ∩ B, is the set containing all elements of A that also belong to B.

The union (denoted by ∪) of a collection of sets is the set of all elements in the collection.

$P(s)$ of a set $s$ is defined as ratio of number of elements in $s$ to the number of elements in $universal set$

given $Universalset=$ $\{$ $\text{2,3,4,5,6,7,8,9,12,13,14,16,20,22,56}$ $\}$

given $A=\{9,12,13,20,22,56\}$ and $H=\{4,5,8,9,16,22\}$

and $C=\{1,4,20,22,56\}$

For Question A:

$A$ ∩ $H$ = $\{9,12,13,20,22,56\}$ ∩ $\{4,5,8,9,16,22\}$

= $\{9,22}\}$

( $A$ ∩ $H$ )∪ $C$ = $\{9,22}\}$ ∪ $C=\{1,4,20,22,56\}$ = $\{1,4,9,20,22,56\}$

$p((A$ ∩ $H)$ ∪ $C)$ = $\frac{6}{15}=\frac{2}{5}$

For Question B:

$H$ ∩ $C$ = $\{4,22\}$

$p(H$ ∩ $C)$ '= $\frac{15-2}{15}=\frac{13}{15}$

For Question C:

$p(H)$ '= $\frac{15-6}{15}=\frac{3}{5}$

For Question D:

$C$ \ $H$ = $\{1,20,56\}$

$p(C$ \ $H)$ = $\frac{3}{15}=\frac{1}{5}$

For Question E:

$A$ \ $C$ = $\{9,12,13\}$

$p(A$ \ $C)$ = $\frac{3}{15}=\frac{1}{5}$

7 0

3 years ago

Solve by elimination 5x-14y=22 -6x+7y=3

Airida [17]

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 -20 - 14y = 22
 + 20 + 20
 -14y = 42
 -14 -14
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 (x, y) = (-4, -3)

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3 years ago

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Gemiola [76]

Answer:

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Step by step explanation:

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2 years ago

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