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Cloud [144]
3 years ago
11

A baker uses 5.5 lb of flour daily how many ounces of flour will he use in two weeks

Mathematics
2 answers:
Shtirlitz [24]3 years ago
7 0
The Baker uses 1,232 ounces in 2 weeks. 
1 pound= 16 ounces.
16 ounces x 5.5 lb= 88
2 weeks=14 days
Now multiply...
88 x 14= 1,232 ounces
Alenkasestr [34]3 years ago
3 0

1 pound = 16 ounces

16 * 5.5 = 88 ounces per day

88*7 = 616 ounces per week

616 * 2 = 1232 ounces in 2 weeks

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The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April,
Karolina [17]

Answer:

a)Null hypothesis:\mu = 24.57  

Alternative hypothesis:\mu \neq 24.57  

b) df=n-1=30-1=29  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(29)}  

c)  If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d)  For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

t_{crit}=\pm 2.045

And the rejection zone would be given by:

(-\infty , -2.045) U(2.045,\infty)

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.

Step-by-step explanation:

The mean hourly wage for employees in goods-producing industries is currently $24.57. Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

a. State the null and alternative hypotheses we should use to test whether the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries.

We need to conduct a hypothesis in order to check if the mean is equal to 24.57 or not, the system of hypothesis would be:  

Null hypothesis:\mu = 24.57  

Alternative hypothesis:\mu \neq 24.57  

b. Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value.

\bar X=23.89 represent the sample mean  

s=2.4 represent the sample standard deviation

n=30 sample size  

\mu_o =24.57 represent the value that we want to test

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

If we analyze the size for the sample is = 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{23.89-24.57}{\frac{2.4}{\sqrt{30}}}=-1.552    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=30-1=29  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(29)}  

c. With  α = 0.05  as the level of significance, what is your conclusion?

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d. Repeat the preceding hypothesis test using the critical value approach.

For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

t_{crit}=\pm 2.045

And the rejection zone would be given by:

(-\infty , -2.045) U(2.045,\infty)

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.

4 0
2 years ago
Mr lee's homeroom class won a pizza party withfive pizzas. The students ate 3 1/3 of the pizzas. How many pizzas are left?
algol13
A whole pizza has 3 x 1/3 pizzas.
So if you eat 3 of 5, 2 pizza lefts.
If you eat 1/3 from 2 pizzas, 1 and 2/3 pizzas left.

Hope this help. :)
3 0
3 years ago
Read 2 more answers
Question is in the picture, please include step-by-step process.
Lemur [1.5K]
You would do the ((2 x 3.14) divided by four) x 3 To get the answer I will explain it if it’s right
4 0
2 years ago
Read 2 more answers
5/7+7/22 what is the answer
andrey2020 [161]

Answer:

1 5/154

Step-by-step explanation:

3 0
3 years ago
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
zvonat [6]

The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0
2 years ago
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