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Serggg [28]
4 years ago
8

What is the value of G? Round only your answer to the nearest tenth.

Mathematics
1 answer:
Anvisha [2.4K]4 years ago
5 0
For this problem we need the definition of law of sines
\frac{ \sin(a) }{a}  =  \frac{ \sin(b) }{b}  =  \frac{ \sin(c) }{c}
therefore we can evaluate our knowns:

<G =74° with side unknown g
<H=58° with side 54

so we can set up the proportion
\frac{ \sin(74) }{g}  =  \frac{ \sin(58) }{54}
then cross multiply and get the value of g

g \times  \sin(58)  = 54 \times  \sin(74)
therefore we get the value of g as

61.2 \: yards

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Question:

What is the answer that you want to find? Fraction to decimal, I get it but what is the fraction? I'm sorry but I didn't get that clear.

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2) Jose recently saw an ad online for a new version of his tablet. The projected price is $450, and the phone will be out on the
soldi70 [24.7K]

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462.38

Step-by-step explanation:

bc its only a year and it is inflated by 2.75% xoxox

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Write the slope-intercept form of the line with a slope of 2 and a y-intercept of -4. Include your work in your final answer. Ty
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Y=mx+b m=slope b=y intercept
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3 years ago
Read 2 more answers
⚡️⚡️Really need help, would appreciate it⚡️⚡️
pentagon [3]

Answer:

The answer to your question is letter A. Regular size is cheaper than Economy size by $0.0003 per gram

Step-by-step explanation:

Cost of 1 gr of economy size

  5000 gr (5 kg) -------------------  $ 5.15

         1 gr            --------------------     X

x = 1 x 5.15 / 5000 = $0.0010

Cost of 1 gr of regular size

  820 gr -------------------   $0.60

       1 gr --------------------   x

x = 1 x 0.6 / 820 = $ 0.0007

Difference of cost  = $0.0010 - $0.0007 = $0.0003

3 0
3 years ago
Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0
3 years ago
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