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Anastaziya [24]
3 years ago
14

David charges $17 plus $7.00 per hour to mow lawns. Ari charges $11 plus $7.75 per hour to mow lawns. In what situations is Ari'

s charge greater than or equal to David's charge?
*PLS HELP*
Mathematics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

Ari's is less than Davids

Ari=18.75

David=24.00

Step-by-step explanation:

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nignag [31]

Answer:

  • 3:16 P.M.
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Step-by-step explanation:

The two dump intervals have a greatest common factor (GCF) of 3, so their least common multiple (LCM) is ...

  (18)(21)/3 = 126 . . . . minutes

This period is 2 hours 6 minutes. The last time both dumped was 1:10, so the next time both will dump is ...

  1:10 +2:06 = 3:16 . . . P.M.

and the next time after that is ...

  3:16 +2:06 = 5:22 . . . P.M.

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3 years ago
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SCORPION-xisa [38]
The correct answer is B, sorry if it wasn't correct.
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Calculate the slope of the line which passes through the points (8,7) and (11, 3)
shepuryov [24]

Answer:

negative 1 1/3

Step-by-step explanation:

you do y2 minus y1 over x2 minus x1 then you simplify.

8 0
3 years ago
Lin rode a bike 20 miles in 150 minutes. If she rode at a constant speed, how far did she ride in 15 minutes? How long did it ta
coldgirl [10]

Hey there! :)

Answer:

First part: 2 miles.

Second part: 45 minutes

Third part: 8 mph.

Step-by-step explanation:

We can divide this question into 3 parts.

Begin by solving for the distance traveled after 15 minutes by creating a ratio:

\frac{20 mi}{150min}  = \frac{x}{15 min}

Cross multiply:

20 · 15 = 150 · x

300 = 150x

300/150 = 150x/150

x = 2 miles.

2nd part: How long it took her to ride 6 miles.

Set up another ratio similar to the one used before:

\frac{20 mi}{150min}  = \frac{6mi}{x min}

Cross multiply:

20 · x = 6 · 150

20x = 900

20x/20 = 900/20

x = 45 minutes.

3rd part:

For this part, we will need to convert from minutes to hours.

\frac{20 mi}{150 min}* \frac{60 min}{1 hr} = \frac{1200mi}{150hr}   = 8 mi/h

Therefore, her speed is 8 mph.

5 0
3 years ago
A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has
ExtremeBDS [4]

Answer:

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

p_v =P(t_{(23)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=136 represent the sample mean

s=11 represent the sample standard deviation

n=24 sample size  

\mu_o =140 represent the value that we want to test

\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 is less than 140, the system of hypothesis would be:  

Null hypothesis:\mu \geq 140  

Alternative hypothesis:\mu < 140  

If we analyze the size for the sample is < 30 and 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{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

P-value

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

df=n-1=24-1=23  

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

p_v =P(t_{(23)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

4 0
3 years ago
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