If it takes 10 machines 10 minutes to make 10 widgets, how long would it take 50 machines to make 50 widgets? minutes

Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is betwe

Dimas [21]

Answer:

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

Step-by-step explanation:

We know mean u = 38 standard dev. s = 2

We want P ( 37 < x < 41)

so

P( (37 - 38) / 2 < Z) = P(-0.5 < Z)

P( Z < (41 - 38)/2 ) = P( Z < 1.5)

Find P(Z < -0.5) = 0.3085

Find P(Z > 1.5) = 0.0668

so P(-0.5 < Z < 1.5) = 1 - P(Z < -0.5) - P(Z > 1.5)

P(-0.5 < Z < 1.5) = 1 - 0.3085 - 0.0668

P(-0.5 < Z < 1.5) = 0.6247

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

3 0

3 years ago

Please Help me plzzzzz<br>click above

aliina [53]

Answer:

google it

..................

5 0

3 years ago

I JUST NEED 4 PLS!!!!!!

Alchen [17]

Answer:

C

Step-by-step explanation:

When doing 4 divided by 2 you get 2. Looking at the sign it is saying 3 is greater than the number we solve for. Since we solved and got 2 we can double check and ask is 3 greater than 2? Yes it is so that would make the statement correct.

Hope this Helps :)

6 0

3 years ago

Find the rectangular coordinates of the point (sqrt3,pi/6)

madreJ [45]

Answer:

$(x, y) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$

Step-by-step explanation:

The rectangular coordinates of the point are:

$(x,y) = \left(\sqrt{3}\cdot \cos\frac{\pi}{6}, \sqrt{3}\cdot \sin\frac{\pi}{6}\right)$

$(x, y) = \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$

5 0

3 years ago

Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that more than 85

Lubov Fominskaja [6]

Answer:

The p value for this test is given $p_v = 0.1071$

Since the p value is higher than the significance level given of $\alpha=0.05$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

Step-by-step explanation:

We define the proportion of interest as p who represent the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workers

And we want to anaylze the following system of hypothesis:

Null hypothesis: $p \leq 0.85$

Alternative hypothesis: $p >0.85$

And the statistic for this test is given by:

$z =\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$

The p value can be calculated with this formula:

$p_v = P(z>z_{calc})$

And the p value for this test is given $p_v = 0.1071$

Since the p value is higher than the significance level given of $\alpha=0.05$ we have enough evidence to FAIL to reject the null hypothesis. And we can say that the true proportion of firms in the manufacturing sector still do not offer any child-care benefits to their workersis is not significantly higher than 0.85 or 85% at 5% of significance.

4 0

3 years ago