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Kruka [31]
2 years ago
6

Use the law of cosines to find the length of a

Mathematics
1 answer:
borishaifa [10]2 years ago
4 0

Answer:

Step-by-step explanation:

You don't need the Law of Cosines, the Law of sines if what you need. You can't use the Law of Cosines because in order to find side a, you would need the length of side c and you don't have it. Using the Law of Sines is appropriate, knowing that angle B = 55:

\frac{sin55}{175}=\frac{sin42}{a} and solving for a:

a=\frac{175sin42}{sin55} so

a = 143.0

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1/7 of all students achieved great success in the written exercise in mathematics. 3/7 very good success. 2/7 good success, whil
laila [671]

Answer:

x = 28

Step-by-step explanation:

Let the total number of students = x

x/7 + 3x/7 + 2x/7 + 4 = x If that is all there is.  If it isn't the problem can't be done. Combine like terms.

x(1 + 3 + 2)/7 + 4 = x

x (6/7) + 4 = x                       Subtract 6x/7 from both sides

4 = x - 6x/7

4 = 7x/7 - 6x/7                     Combine

4 = 1/7x                                Multiply both sides by 7

4*7 = 7x / 7

x = 28

6 0
2 years ago
Bella is x years old. Jake is 9 years older than Bella. Five times Jake’s age is equal to 300. (a) Write an equation that could
podryga [215]
300 divided by 5= jake
Jakes answer minus 9 is Bella’s age
3 0
2 years ago
For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co
Rzqust [24]

Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b)   Margin of error ( E) = $1,000 , n = 216

(c)   Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation \sigma = $7,500

\alpha = 1 - confidence interval = 1 - .95 = .05

Z_{\frac{\alpha}{2}} =  Z_{\frac{.05}{2}} = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E)  = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

                           E   = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}

n =  54.0225

n = 54 ( approximately)

(b)   Margin of error ( E) = $1,000

          E     = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

         1000   =  Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}

n = 216

(c)   Margin of error ( E) = $500

   E = Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}

  500 = Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}

Squaring both side

500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}

n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}

n = 864

7 0
3 years ago
8. Which expression is greatest? Justify your answer.<br> a. 4 X -5<br> b. -5 + 4<br> c. -4- (-5)
vladimir1956 [14]

Answer:

c

Step-by-step explanation:

i think this is the greatest because if you solve all of the expressions you will find out that this is the greatest.

a.4×-5=-20

b.-5+4=-1

c.-4-(-5)=1

therefore 1 is the greatest so the expression c is the greatest.

I hope this helps

3 0
3 years ago
Read 2 more answers
PLEASE HELP! NO LINKS AND DONT ANSWER JUST FOR POINTS OR U WILL BE REPORTED! Will give brainiest!!
chubhunter [2.5K]

Answer:

brainly.com

Step-by-step explanation:

sorry my bad copy paste error

5 0
2 years ago
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