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masya89 [10]
3 years ago
7

A paper company produces 1,740 notebooks in 2 days. How many notebooks can it produce in 10 days?

Mathematics
1 answer:
vladimir1956 [14]3 years ago
8 0

Answer: in 10 days the company can make 87000 notebooks considering the company produces at the same rate.


Step-by-step explanation:

Think 1740=2d

X=10d

Cross multiply

2x= 17400

X= 8700

8700 x 10d = 87000

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(3+3+8-9)+(66-6)6=<br><br><br> Please give correct answer if you a brainlest
yanalaym [24]

Answer:

365

Step-by-step explanation:

3+3=6+8=14-9=5+360=365                                                                                                                  

                              /\

                           66-6=60*6=360

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2 years ago
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Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa
xxMikexx [17]

Answer:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      32   18    50

Medium                                 68     7    75

Large                                     89    11    100

_____________________________________

Total                                     189    36   225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1:  NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{50*189}{225}=42

E_{2} =\frac{50*36}{225}=8

E_{3} =\frac{75*189}{225}=63

E_{4} =\frac{75*36}{225}=12

E_{5} =\frac{100*189}{225}=84

E_{6} =\frac{100*36}{225}=16

And the expected values are given by:

Size Company/ Heal. Ins.   Yes   No  Total

Small                                      42    8    50

Medium                                 63     12    75

Large                                     84    16    100

_____________________________________

Total                                     189    36   225

And now we can calculate the statistic:

\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(3-1)(2-1)=2

And we can calculate the p value given by:

p_v = P(\chi^2_{2} >19.221)=0.000067

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example \alpha=0.05, we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

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Answer: adjoining HAVE A GREAT DAY

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According to the National Association of Colleges and Employers, the average starting salary for new college graduates in health
katen-ka-za [31]

Answer:

a) The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

b) The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

c) The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

d) A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

Step-by-step explanation:

<em>a. What is the probability that a new college graduate in business will earn a starting salary of at least $65,000?</em>

For college graduates in business, the salary distributes normally with mean salary of $53,901 and standard deviation of $15,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{65000-53901}{15000} =0.74

The probability is then

P(X>65,000)=P(z>0.74)=0.22965

The probability that a new college graduate in business will earn a starting salary of at least $65,000 is P=0.22965 or 23%.

<em>b. What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000?</em>

<em />

For college graduates in health sciences, the salary distributes normally with mean salary of $51,541 and standard deviation of $11,000.

To calculate the probability of earning at least $65,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{65000-51541}{11000} =1.22

The probability is then

P(X>65,000)=P(z>1.22)=0.11123

The probability that a new college graduate in health sciences will earn a starting salary of at least $65,000 is P=0.11123 or 11%.

<em>c. What is the probability that a new college graduate in health sciences will earn a starting salary less than $40,000?</em>

To calculate the probability of earning less than $40,000, we can calculate the z-value:

z=\frac{x-\mu}{\sigma} =\frac{40000-51541}{11000} =-1.05

The probability is then

P(X

The probability that a new college graduate in health sciences will earn a starting salary of less than $40,000 is P=0.14686 or 15%.

<em />

<em>d. How much would a new college graduate in business have to earn in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences?</em>

The z-value for the 1% higher salaries (P>0.99) is z=2.3265.

The cut-off salary for this z-value can be calculated as:

X=\mu+z*\sigma=51,541+2.3265*11,000=51,541+25,592=77,133

A new college graduate in business have to earn at least $77,133 in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences.

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Answer:

The last one

Step-by-step explanation:

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