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stira [4]
3 years ago
7

a machine can produce 250 items per hour. if the operator of the machine works 40 hours per week, how many times are made by thi

s machine in one year?
Mathematics
2 answers:
Harman [31]3 years ago
6 0
The machine would produce 520,000 in one year
pantera1 [17]3 years ago
3 0
Assuming that the year has 52 weeks in it, (not a leap year) the machine would produce 520,000. You need to multiply the amount of hours worked by how many items the machine makes in an hour. 40x250=10000. Then you need to multiply the amount of items made in a week by 52 (the number of weeks in a year). 10000x52=520000.
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Create an expression that represents the sum of 9.5 and n
AlladinOne [14]

Answer:

Step-by-step explanation:

9.5 + n

n + 9.5 is also allowed.

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Bobby bought 3 5/12 ft of red ribbon and 5 1/12 ft of yellow ribbon.
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The answer is B) $10.20

8  \frac{6}{12} \:   is \: same \: as  \: \: 8 \frac{1}{2}

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3 years ago
I need help with average rate of change please help ASAP
babunello [35]
When you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. f(x) = x2 and f(x + h) = (x + h) Therefore, the slope of the secant line between any two points on this function is 2x + h.
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4 years ago
A recent study focused on the number of times men and women send a WeChat message in a day. The information is summarized next.
Ratling [72]

Answer:

1)Null hypothesis:\mu_{men}=\mu_{women}

Alternative hypothesis:\mu_{men} \neq \mu_{women}

2) Two critical values are z_{\alpha/2}=-2.58 and z_{1-\alpha/2}=2.58

3) z=-2.40  

4) p_v =2*P(z

5) Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.

Step-by-step explanation:

Data given and notation

\bar X_{men}=23 represent the mean for the sample men

\bar X_{women}=28 represent the mean for the sample women

\sigma_{men}=5 represent the population standard deviation for the sample men

\sigma_{women}=10 represent the population standard deviation for the sample women

n_{men}=25 sample size for the group men

n_{women}=30 sample size for the group women

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

1)Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{men}=\mu_{women}

Alternative hypothesis:\mu_{men} \neq \mu_{women}

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

z=\frac{\bar X_{men}-\bar X_{women}}{\sqrt{\frac{\sigma^2_{men}}{n_{men}}+\frac{\sigma^2_{women}}{n_{women}}}} (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

2)Determine the critical value(s).

Based on the significance level\alpha=0.01 and \alpha/2=0.005 we can find the critical values with the normal standard distribution, we are looking for values that accumulates 0.005 of the area on each tail on the normal distribution.

For this case the two values are z_{\alpha/2}=-2.58 and z_{1-\alpha/2}=2.58

3) Calculate the value of the test statistic for this hypothesis testing.

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

z=\frac{23-28}{\sqrt{\frac{5^2}{25}+\frac{10^2}{30}}}}=-2.40  

4)What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

p_v =2*P(z

5)Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the mean number of times men and women send a Twitter message in a day.

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4 0
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