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Alika [10]
3 years ago
6

What is 5 2/3 in radical form?

Mathematics
1 answer:
velikii [3]3 years ago
6 0
5^(2/3)=25^(1/3)
Third one ( the one below your selection)
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If cos ø = -4/7 what are the values of sin ø and tan ø
densk [106]

Answer:

tan=5,5

sin=0,98

Sry when Its wrong

6 0
3 years ago
Read 2 more answers
In science class sara needed 8 test tubes for 3 different experiments. The first experiment required 2 test tubes and the other
photoshop1234 [79]

Answer:

as we know Sara need 8 tube in 3 experiment

she will use 2 tube in 1 experiment

now she have 6 tube in other experiments

so the question said to find how many tube were used in both experiment equally so 3,3 test tube were needed for each of the other two experiment

Step-by-step explanation:

total tube=8 (3 experiment)

used =2tube (1 experiment)

remaining=8-2=6

now ,

dividing 6 into 2 equal part

so,=6/2=3

so 3 tube were used in 2nd experiment and 3in 3rd experiment

6 0
3 years ago
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, T
myrzilka [38]

Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5                          

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

\text{P(at least one head)} = \dfrac{7}{8} = 0.875

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5

3 0
3 years ago
Question 12 down at the bottom of the file PDF<br><br> Im confuse on question 12
mezya [45]
Nothing shows up.. could you type it out?
7 0
3 years ago
Need help doing this question, i think i got all of them wrong. If you explain your thought process it'd be nice too!
icang [17]

Answer:

4 with 3

Step-by-step explanation:

the only ones that are perpendicular to each other are

y = 7/4 x - 2 with y = - 4/7 + 2

since the condition is met so that two lines are perpendicular

slope1 * slope2 = -1

7/4 * -4/7 = -1  correct

8 0
3 years ago
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