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Travka [436]
3 years ago
15

A delivery person uses a service elevator to bring boxes of books up to an office. The delivery person weighs 190 lb and each bo

x of books weighs 60 lb. The maximum capacity of the elevator is 1740 lb. How many boxes of books can the delivery person bring up at one​ time?
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
6 0

Answer:

25 boxes.

Step-by-step explanation:

We have the inequality:

190 + 60x ≤ 1740  where  x = the number of boxes of books.

60x  ≤  1740 - 190

60x   ≤  1550

x ≤ 25.83.

aksik [14]3 years ago
5 0

Answer:

25 boxes

Step-by-step explanation:

To get the amount of boxes of books that can be brought up, subtract the weight of the delivery person from the maximum capacity: 1740 - 190 = 1550 lbs left to fill by the boxes

To find the amount of books able to be brought up, divide the capacity left by the weight of each of the boxes: 1550/60 = 25.8 books able to be carried up, without exceeding, meaning that the delivery person can bring up 25 boxes of books without the service elevator collapsing

Hope this helps :)

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In which number does the digit 7 have a value that is 10 times as great as the digit 7 in the number 0.975
Inessa05 [86]

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5 0
3 years ago
he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ > 15 has
DedPeter [7]

Answer:

t = 2.24

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

df=n-1=23-1=22  

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

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

Step-by-step explanation:

Data given and notation  

\bar X represent the mean height for the sample  

s represent the sample standard deviation

n=23 sample size  

\mu_o =15 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

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

For this case the statistic is given:

t = 2.24

P-value

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

df=n-1=23-1=22  

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

p_v =P(t_{(22)}>2.24)=0.01776  

And for this case we can conclude that:

0.01 < p_v < 0.025

And we will reject the null hypothesis at \alpha=0.025 since p_v < \alpha

5 0
3 years ago
Suppose that 11 inches of wire costs 66 cents. at the same rate, how much (in cents) will 4 inches of wire cost?
Andre45 [30]
24 cents.

11/66= 6
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(proportions)

hope this is clear :))
7 0
3 years ago
Mark was thinking of a number. Mark doubles it, then adds 11 to get an answer of 52.6. What was the original number?
Artemon [7]

Answer:

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Step-by-step explanation:

The first step you need to do is to subtract 11:

52.6 - 11 = 41.6

The second step is to divide it by 2:

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Mark was thinking of 20.8

5 0
2 years ago
Read 2 more answers
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