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Firdavs [7]
3 years ago
13

To approximate the length of a marsh, a surveyor walks x=420 meters from point A to point B,then turns 75 degrees and walks 220

meters to point C (see figure). Approximate the length AC of the marsh. (Round your answer to one decimal place.)
Mathematics
1 answer:
enot [183]3 years ago
6 0

To approximate the length of a marsh, a surveyor walks x=420 meters from point A to point B,then turns 75 degrees and walks 220 meters to point C (see figure). Approximate the length AC of the marsh. (Round your answer to one decimal place.)

<u>37.1 meters.</u>

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15 feet of edging costs $24.98. How much will 28 yards cost?
GuDViN [60]

Answer:

  139.888 $

Step-by-step explanation:

Data :

   15 ft = 24.98$

   So

  1 yard = 3 ft

15ft / 3 ft = 5 yard

So  

   5 yard  = 24.98$

28 yd / 5yd = 5.6

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Find the lest number which is exactly divisible by 36 &amp; 90 Maths ​
prisoha [69]

Answer:

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Step-by-step explanation:

You need to find the HCF of 36 and 90.

<u>Prime factors of each:</u>

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<u>HCF includes all prime factors of both numbers:</u>

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3 0
3 years ago
Read 2 more answers
A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
Gnom [1K]

Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

The degrees of freedom are given by:

df=n-1=93-1=92  

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

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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Answer:

Your Answer is B.

Step-by-step explanation:

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