*PLEASE HELP ME!!!!!**

2 answers:

6 0

For the first question, the answer is A. You take 90 and multiply it by 3, then you get the time that it takes for one worker to mow the large park.

For the second question, you take 50 = k/0.25. To find k, you would want to multiply both sides with 0.25 to cancel the 0.25 on k's side. 50 times 0.25 is 12.5. I think C is your answer although I'm not really confident.

pentagon [3]3 years ago

4 0

Answer:

For the second one the answer is 6.25 I just took the quiz

Step-by-step explanation:

Hope it helps:)

You might be interested in

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$