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N76 [4]
3 years ago
11

This is a practice asaigment before the real assignment plz help​

Mathematics
1 answer:
Annette [7]3 years ago
3 0

Answer:

1)a+$5.92=$12.29

a= $12.29-$5.92

a=$6.37

2) 42/j = 6

6j=42

j=42/6

j=7

3) j+8=72 —> 9+8=72–>17=72 False

m x 3= 27 —> 9x3=27 —> 27=27 True

b+4=13 —> 9+4=13 —> 13=13 True

p-5=4 —> 9-5=4 —> 4=4 True

q/3 = 27 —> 9/3 =27 —> 3=27 False

4) $15.32+ C= $29.00

For soccer ball: C= $12.87

$15.32+ C= $29.00 —> $15.32+ $12.87 = $29.00 —> $28.19

For baseball cap: C=$8.39

$15.32+ C= $29.00 —> $15.32+ $8.39 = $29.00 —>$23.71

For a set of shin guard: C=$14.98

$15.32+ C= $29.00 —> $15.32+ $14.98 = $29.00 —>$30.3

For water bottle: C=$5.93

$15.32+ C= $29.00 —> $15.32+ 5.93 = $29.00 —>$21.25

Because annabetg can spend only $29 at the store, the most expensive item she can buy with the t-shirt is the soccer ball which cost $12.87

5) 6x=12–> x=12/6 —> x=2 False

5+x=11 —> x=11-5 —> x=6 True

x/2=3 —> x= 2x3 —> x=6 True

17-x=11 —> -x=11-17 —> -x=-6 —> x=6 True

Note: Review part 4 please or ask someone to resolve it to make sure it is correct

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Answer:

Therefore, Steve will paint house for 20.6 days.

Step-by-step explanation:

We know that Steve determines that it would take them 9 days to paint the house together (if they were both healthy) and that it would take Janet (when healthy) 16 days to paint the house alone.

We conclude that in 1 day Janet painted 1/16 of the house. In nine days, Janet painted 9/16 of house.

So, 7/16 of house paint Steve in nine days.

We have the following proportion:

\frac{7}{16}:9=1:x\\\\\frac{7}{16}x=9\\\\x=9\cdot \frac{16}{7}\\\\x=20.6

Therefore, Steve will paint house for 20.6 days.

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Company a= 15x+10y+25z+75

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Step-by-step explanation:

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\\ \tt\longmapsto 9(a+11)

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Consider a single spin of the spinner.
alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed.  These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0
3 years ago
Read 2 more answers
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