Choose 1 answer:

PLZ LOOK AT THE PICTURE AND HELP ME. ALSO EXPLAIN HOW U GOT THE ANSWER

Sati [7]

If it's found at -331 degrees it would be 331 degrees below 0 because 0-331=-331. So it would be the second option.

8 0

3 years ago

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 200, \sigma = \sqrt{400} = 20$

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{150 - 200}{20}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So $Z = 1.28$

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 200}{20}$

$X - 200 = 20*1.28$

$X = 225.6$

225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

6 0

3 years ago

Provolone cheese costs $5.00 per pound. We can get 6 slices per pound. You need 8 slices for your recipe. How much does the chee

Grace [21]

Answer:

6 dollars and 67 cents

Step-by-step explanation:

4 0

2 years ago

The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si

natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b) y = 1.148*x + 6.036

Step-by-step explanation:

Given:

f(x) = 6 - 10*x^2

g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

f'(x) = -20*x

g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

m = (g(x_o) - f(x)) / (x_o - x)

m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

m = f'(x) = g'(x_o)

- We will develop the first expression:

m = f'(x)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

m = g'(x_o)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

(x - x_o)(20*x - 2*x_o + 4) = 0

x = x_o , x_o = 10x + 2

- For x_o = 10x + 2 ,

(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

(-90*x^2 + 2) = -180*x^2 - 40*x

90*x^2 + 40*x + 2 = 0

- Solve the quadratic equation above:

x = -0.0574, -0.387

- Largest slope is at x = -0.387 where equation of line is:

y - 4.502 = -20*(-0.387)*(x + 0.387)

y = 7.74*x + 7.5

- Other tangent line:

y - 5.97 = 1.148*(x + 0.0574)

y = 1.148*x + 6.036

6 0

3 years ago

Helpppppp!

Dovator [93]

Answe The locations of E' and F' are E' (−8, 0) and F' (0, 4), and lines g and g' intersect at point F.

The locations of E' and F' are E' (−4, 0) and F' (0, 2), and lines g and g' are the same line.

The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.

The locations of E' and F' are E' (−1, 0) and F' (0, 0), and lines g and g' are not related.

are your answer options I went with.. The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.

Step-by-step explanation:

5 0

2 years ago