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Stells [14]
3 years ago
14

Only one answer is correct help

Mathematics
2 answers:
kupik [55]3 years ago
5 0
C is correct ! did this help
alexdok [17]3 years ago
3 0

Answer:

9/8 - 2/3 = 11/24

C is the correct answer.

Let me know if this helps!

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Linda shoots an arrow at a target in an archery competition. The arc of the arrow can be modeled by the equation y= -0.02x to th
rusak2 [61]

Answer:

37.8metres

Step-by-step explanation:

The arc of the arrow can be modeled by the equation:

y=-0.02x²+0.65x+4

Where x is the horizontal distance (in meters) from Linda and y is the height (in meters) of the arrow.

The arrow hits the ground when its height (y) is zero.

Therefore, we determine the value(s) of x for which:

y=-0.02x²+0.65x+4=0

Using a calculator to solve the quadratic equation:

x=37.79 or -5.29

Since the distance cannot be a negative value, we ignore -5.29.

The distance from Linda when the arrow hits the ground is 37.8metres (to the nearest tenth)

7 0
3 years ago
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I'm stuck on this question, PLZ HELP. --Mama Mia restaurant used 1/2 of their mozzarella cheese making pizza and the remaining 6
vivado [14]
If Mama Mia used half of their mozarella cheese, that means that 64 ounces is the other half. 64 times 2 equals 128.
6 0
4 years ago
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Calculate the mediamnumber of hits which player listed below had the same number of hits as the median?
ch4aika [34]

Answer:

you have to finish the question

Step-by-step explanation:

6 0
3 years ago
What is the answer to this?
Nonamiya [84]

Answer:

27

Step-by-step explanation:

90-63

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3 years ago
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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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