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Gala2k [10]
3 years ago
8

a new energy drink advertise 182 calories for 6 ounces how many calories are in 21 ounces of the drink? There are ____ calories

in the 21 ounce drink
Mathematics
2 answers:
pantera1 [17]3 years ago
5 0

Answer:

636.9 calories = 21 ounces

Step-by-step explanation:

182 calories= 6 ounces

30.3 calories= 1 ounces

636.9 calories= 21 ounces

liberstina [14]3 years ago
4 0
I think it should be 637
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P: It's sunny.
lapo4ka [179]

Answer:

I think the answer is c

Step-by-step explanation:

please give me brainlist

5 0
3 years ago
58 yd<br> 73 yd<br> Use 3.14 Instead Of Pie Symbol
defon

Answer:

771096.8 \: yd^{3}

Step-by-step explanation:

If you are trying to find the volume, you would use the formula:

\pi \: r ^{2} (h)

You would plug in the radius (r), which is 58, and then plug in the height, which is 73.

Make sure you remember to use 3.14 and not the pi key if you are using a calculator.

The final answer is 771,096.8 yd^3

7 0
3 years ago
I’m confused on this question. Can anybody help?
DENIUS [597]

Answer:

I believe the answer is 90

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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
The histogram shows the speeds of downhill skiers during a competition. Estimate the mean of the data set displayed in the histo
Ostrovityanka [42]

Answer:

≈50.6

Step-by-step explanation:

Not sure what precision level this problem is looking for, but for right-skewed distributions, we know that the mean is going to be pulled right and therefore the mean should be larger than the median. To a high confidence level, the mean should fall between 50 and 59, or in the third column.

If a single estimation is wanted, assume the values inside each column are uniformly distributed:

\displaystyle\\\widehat{\mu}=\frac{34.5\cdot 4+...+74.5\cdot 2}{4+9+4+4+2}\approx \boxed{50.6}

8 0
3 years ago
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