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Galina-37 [17]
3 years ago
7

Evaluate ^3 sqrt 0.008- need answer asap!!

Mathematics
1 answer:
Ann [662]3 years ago
7 0
∛0.008 = ∛(0.2)^3
= 0.2

hope it helps
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Part C
charle [14.2K]

Answer:

(x,y+7)

Step-by-step explanation:

A translation means you are moving the image a certain amount of units, in this case, 7 units up. Since x values are all horizontal x does not change, but all y values will be 7 units higher.

For example (1,2) becomes (1,9).

8 0
3 years ago
Solve<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3x%20%2B%204%7D%7B2%7D%20%20%3D%209.5" id="TexFormula1" title=" \frac{3x
ELEN [110]
Answer: 5

Step by step:

(3x+4)/2 = 9.5
3x+4 = 9.5*2
3x+4 = 19
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7 0
3 years ago
Read 2 more answers
Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac
valentinak56 [21]

Answer:

tan(θ) = 0, 0.577, -0.577

Step-by-step explanation:

3tan³(θ) - tan(θ) = 0

tan(θ)(3tan²(θ) - 1) = 0

tan(θ) = 0

tan²(θ) = ⅓ tan(θ) = +/- sqrt(⅓)

tan(θ) = 0, sqrt(⅓), -sqrt(⅓)

tan(θ) = 0, 0.577, -0.577

To find θ values, domain is required

6 0
3 years ago
Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar
horsena [70]

Answer:

P(A) = \frac{1}{9}

P(B) = \frac{1}{3}

P(C) = \frac{1}{729}

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

P(M) = \frac{1}{3}.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is P(A) = \frac{1}{9}. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}.

C. If the probability that two people have been hired on a specific day of the week is 3(\frac{1}{3}) ^ 2, then the probability that 7 people have been hired on the same day is:

P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}

D. The probability is \frac{1}{729}. This number is quite close to zero. Therefore it is an unlikely bastate event.

6 0
3 years ago
Show all the nesessary steps​
inessss [21]

what's up? to convert this into a fraction, simply multiply 12.3/1 by 10 in order to get 123/10. you can also do it by seeing .3 is in tenths, so it would be 3/10. you are left with 12 (whole number) so 12 3/10 which is also 123/10

best of luck with your studies

4 0
3 years ago
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