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Orlov [11]
3 years ago
10

What statement is true? (Help needed)

Mathematics
1 answer:
Goshia [24]3 years ago
4 0
B the correct answer Is b
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A store manager combines 2 lb of trail mix that contains some cashew nuts with 30 lb
bekas [8.4K]

Answer:

100%

Step-by-step explanation:

Let  p = percent of cashews in first mix

2p = lb. of cashews in first mix

30(.20) =  6 = lb. of cashews in second mix

32(.25) =  8 = lb. of cashews in the combined mix

So,

lb. of cashews in first mix +  lb. of cashews in second mix = lb. of cashews in the combined mix

2p + 6 = 8

2p = 2

p = 1 = 100%

7 0
2 years ago
Read 2 more answers
A serving of Ice cream contains 6400 calories. 1600 calories come from fat. What percent of the total calories come from fat
geniusboy [140]
Just divide.

1600 / 6400 = 0.25

0.25 * 100 = 25%
7 0
3 years ago
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Find the slope of the line that passes through the pair of points.(1,7) (10,1)
VladimirAG [237]
So I believe you go
( y_{1} - y_{2} )                  (x_{1}- x_{2)}

and substitute x an y like this 
  
(1-7)         (7-1)
and get a slope of -9/6 

Sorry If I got it wrong,but I hope I helped
7 0
3 years ago
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4.) The graph below shows that the cost of bananas depends on the number of pounds
Alina [70]

Answer: B.$1.25

Step-by-step explanation: 10 divided by 1.25= 8 pounds

8 0
3 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.

3 0
3 years ago
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