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2 years ago

Solve 7e^(2x)-5>58. Round to the nearest hundredth.

Mathematics

1 answer:

7nadin3 [17]2 years ago

5 0

Answer:

$x\gtrapprox1.1$

Step-by-step explanation:

$7e^{2x}-5>58$

$7e^{2x}>63$

$e^{2x}>9$

$2x>ln(9)$

$x>\frac{ln(9)}{2}$

$x\gtrapprox1.1$

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3 years ago

Javier asks his mother how old a tree in their yard is. His mother says, “The sum of 10 and two-thirds of that tree’s age, in ye

sdas [7]

Answer:

$10 + (\frac{2}{3}) a = 50$ is the CORRECT equation.

Step-by-step explanation:

The given question is INCOMPLETE.

Javier asks his mother how old a tree in their yard is. His mother says, “The sum of 10 and two-thirds of that tree’s age, in years, is equal to 50.” Javier writes the equation { 10 + 2/3} where a is the tree’s age in years. His equation is not correct. What error did he make?

Now here:

a: The tree’s age in years.

Also, “The sum of 10 and two-thirds of that tree’s age, in years, is equal to 50.”

⇒ 10 + two-thirds of that tree’s age = 50

$\implies 10 + (\frac{2}{3}) a = 50$

But in the equation written by Javier, the the third fraction is NOT MULTIPLIED by the age of the tree a in Years.

So, the written equation by Javier is Incorrect.

Now, solving the written correct equation for the value of a, we get:

$\implies 10 + (\frac{2}{3}) a = 50\\\implies (\frac{2}{3}) a = 40\\\implies a = 40 \times (\frac{3}{2}) = 60\\\implies a = 60$

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3 years ago

Find the probability that the mean annual preciptiation will be between 32 and 34 inches. variable is normally distributed

liq [111]

Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.

This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
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Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.

My TI-83 calculator provided the following result:

normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)

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3 years ago