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lisov135 [29]
3 years ago
7

Catarina has 57 gumballs. Catarina keeps 1 gumballs. Catarina has 7 friends and gives the remaining gumballs to them. Each frien

d gets the same amount of gumballs. How many gumballs does each friend get?
Mathematics
1 answer:
DedPeter [7]3 years ago
4 0
Catarina gives each friend 8 gumballs
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If AB=BC,AB=4=4x-2,andBC=3x+3,fine the length of AB
laila [671]

Answer:

18

Step-by-step explanation:

First, I'm assuming AB=4=4x-2 was a typo and it's supposed to be AB = 4x - 2

AB=BC

AB = 4x - 2 BC = 3x + 3

4x - 2 = 3x + 3

Solve for x Add 2 to each side

4x - 2 = 3x + 3

4x - 2 + 2 = 3x + 3 + 2

4x = 3x + 5 Subtract 3x from each side.

4x - 3x = 3x- 3x + 5

4x - 3x = 5

x = 5

Now plug back in to the original equations

AB = 4x - 2                  BC = 3x + 3

AB = 4 (5) - 2               BC = 3(5) + 3

AB = 20 - 2                  BC = 15 + 3

AB = 18                        BC = 18

So AB is 18

3 0
3 years ago
How do I solve the linear equation of 27=6(x)+4(y) over 20=2(x)+5(y)
larisa86 [58]
<u>27 = 6x + 4y </u>= 1 7/20 = 3 4/5<u>
</u>20 = 2x + 5y<u>
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8 0
3 years ago
 Please Help!!!
Alexandra [31]
The answer is 1.715 hope I helped
4 0
3 years ago
In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sa
Deffense [45]

Answer:

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

Step-by-step explanation:

Assuming this complete problem: "The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization's staff of salespersons. He believes that the proportion of women in similar sales positions across the country is less than 45%. Hoping to find support for his belief, he directs you to test

H0: p .45 vs H1: p < .45.

In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women.

Compute the p-value associated with this test. Place your answer, rounded to 4 decimal places, in the blank. For example, 0.3456 would be a legitimate entry."

1) Data given and notation

n=50 represent the random sample taken

X=18 represent the number of women in the sample selected

\hat p=\frac{18}{50}=0.36 estimated proportion of women in the sample

p_o=0.45 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of women is less than 0.45:  

Null hypothesis:p\geq 0.45  

Alternative hypothesis:p < 0.45  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

4 0
3 years ago
A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without
sveticcg [70]

Question:

For the following exercise, determine the range (possible values) of the random variable. A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without replacement, until a nonconforming part is obtained. The random variable is the number of parts selected.

Answer:

Possible values are all integers from 1 to 491.

{1,2,...........491}

Step-by-step explanation:

Given:

Total number of machined parts = 500

Number that do not conform to customers requirements = 10

Number that conforms to customers requirements = 500 - 10 = 490

Since, parts are selected successively, without replacement, until a nonconforming part is obtained, the range of the random variable (number of parts selected) here will be all integers from 1 to 491 ie {1,2,...........,491}.

Here, we have a total of 490 conforming parts, and 10 non conforming parts, there must be a non-conforming part among 491 selections, considering the fact that total number of parts are 500

7 0
3 years ago
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