All categories

2 years ago

Which of the following employees has the largest gross pay?

Mathematics

1 answer:

Leona [35]2 years ago

5 0

Answer:

C po ang Alam kong sagot

Step-by-step explanation:

it's look like a test

You might be interested in

Our library has 3,489 non-fiction books, 8,617 fiction books and 1,240 reference books. If there are 564 students and each stude

Pavlova-9 [17]

Answer:

9,962

Step-by-step explanation:

3,489+8,617+1,240=13,346

564*6=3,384

13,346-3,384=9,962

3 0

3 years ago

Judson collected information about the name length and the population of a random sample of 296 American cities. Here are the re

Natasha2012 [34]

It would be 90 - 10

6 0

3 years ago

Helppppppppppppppppppppppppppppppppppppp

almond37 [142]

i think it would be D

4 0

2 years ago

A bag contains 42 red, 45 green, 20 yellow, and 32 purple candies, you pick one candy at random, find the probability that it is

Gnesinka [82]

Answer:

P (green/yellow) = 65/139

Step-by-step explanation:

42 + 45 + 20 + 32 = 139 possibilities

45 + 20 = 65 desired outcomes

65/139

Hope this helps!

3 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago