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GarryVolchara [31]
3 years ago
13

What is the product in simplest form?

Mathematics
2 answers:
julsineya [31]3 years ago
6 0
When multiple an negative and positive it was always equal negative so -89 times 56 will equal -2027

Fill free to ask anymore question if u need help.
boyakko [2]3 years ago
4 0
The answer to this question is -1315.
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The cost to manufacture a hair clip I $0.50 with a markup of 200 percent what is the selling price of this hair clip
andrey2020 [161]

Answer: i don't rlly no but here is something


Step-by-step explanation:

If the markup on an oven is 200% based on cost, what is the corresponding percent markup based on selling price? Round percent's to the nearest tenth of a percent. Do not enter the percent symbol in your answer.

8 0
3 years ago
Ten digit telephone number has a form (XYZ) - ABC-DEFG. In the particular state there are 4 possible area codes (404, 470, 678,
mars1129 [50]

Answer:

im not sure but i rember having this question i think its 770

Step-by-step explanation:

6 0
3 years ago
2/6<br> Please I need help on this
maria [59]

Answer:

1/3

Step-by-step explanation:

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7 0
3 years ago
1.) Solve for c:<br> 4c + 5 &gt; -c-5
Svet_ta [14]

Answer:

Step-by-step explanation:

5 0
4 years ago
On a multiple choice test, if you randomly guessed on three questions, then what is the probability you got at least one of them
snow_lady [41]

Answer:

Number of Questions =3

Probability of giving a correct answer

                              =\frac{1}{3}

Probability of giving two correct answers

                       =\frac{2}{3}

Probability of giving all correct answers

            =\frac{3}{3}\\\\=1

Probability that at least one of them is correct

         =_{1}^{3}\textrm{C}\\\\=\frac{3!}{(3-1)! \times1!}\\\\=3 \text{ways}\\\\=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\\\=\frac{1}{27}

Probability that two of them is correct

         =_{2}^{3}\textrm{C}\\\\=\frac{3!}{(3-2)! \times2!}\\\\=3 \text{ways}\\\\=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\\\=\frac{8}{27}

Probability that all of them is correct

         =_{3}^{3}\textrm{C}\\\\=\frac{3!}{(3-3)! \times3!}\\\\=1 \text{way}\\\\=1

So, Required probability

         =\frac{1}{27} \times \frac{8}{27} \times 1\\\\=\frac{8}{729}

6 0
3 years ago
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