Which expressions are equivalent to <br> 3x+3(x+y)

A warehouse worker fills 150 orders per day on average. From day to day, the number of orders varies by 2%. What is the range of

choli [55]

Answer:

The range is from 147 to 153 orders per day

Step-by-step explanation:

orders varies by 2% means that orders can be LOWER THAN THE AVERAGE, or HIGHER THAN THE AVERAGE.

That is, by 2%.

First, we need to find the decimal of 2%, so

2/100 = 0.02

We multiply this with the average number of order, 150, to get the varying amount:

0.02 * 150 = 3

Thus, the range would be:

150 - 3 = 147

150 + 3 = 153

The range is from 147 to 153 orders per day

5 0

3 years ago

How to solve this I am not the best at math

Brrunno [24]

Try it is a great site that you can plug problems in to for the answers! I hope this is helpful!!

5 0

3 years ago

Simplify the radical expression below:<br> Square root of 80

marta [7]

Answer:

$4\sqrt{5}$

Step-by-step explanation:

$\sqrt{80}$

$\sqrt{16} \sqrt{5}$

$4\sqrt{5}$

8 0

3 years ago

Read 2 more answers

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-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".

Calculate the statistic

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

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

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

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

3 years ago

I need help with this

Nina [5.8K]

Answer:

Its 40.9!

Step-by-step explanation:

The opposite sides of angles are most of the time congruent, or the same!

Hope this helps and don't forget to follow!

5 0

2 years ago

Which expressions are equivalent to 3x+3(x+y)