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makvit [3.9K]
3 years ago
12

Please help! Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. C = 67°, a = 2

1, c = 20
Mathematics
2 answers:
lys-0071 [83]3 years ago
6 0
(20/Sin 67) = (21/Sin A)
Cross multiply
21Sin67 = 20SinA
Divide by 20
(21Sin67)/20 = SinA

A = Sin^-1 (21Sin67)/20 = 75 degrees
ValentinkaMS [17]3 years ago
3 0

Answer:

A = 75.1°, B = 37.9°, b = 13.3; A = 104.9°, B = 8.1°, b = 3.1


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NEED HELP ASAP! 40 POINTS!
vladimir1956 [14]

Same height when both equations are equal

-16x^2 +  74x + 9 = -16x^2 + 82x

82x - 74x = 9

8x = 9

x = 9/8

x = 1.125

x = 1.13 ---->rounded to nearest hundredths of seconds

answer

1.13 seconds


3 0
4 years ago
Read 2 more answers
A water gauge in a pond measured 2358 in. At the beginning of the week. After five days of rain, the gauge read 3414 in. By how
V125BC [204]

Answer:

1056 inches

Step-by-step explanation:

Given that :

Initial measurement = 2358 in

Guage level after 5 days of rain (final reading) = 3414 in

Change in water level rise :

Final reading - Initial reading

(3414 - 2358) in

= 1056 inches

3 0
3 years ago
Choose the equivalent system of linear equations that will produce the same solution as the one given below.
dmitriy555 [2]
<span>C: 3x + 2y = 6
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6 0
4 years ago
Rewrite the expression as a multiple of a sum of two numbers with no common factor 26 + 22
uysha [10]

2(11+13) = 26+2w because 2(11)=22 2(13)=26

7 0
3 years ago
The branch manager of an outlet (Store 1) of a nationwide
monitta

Answer:

1. ( 19.1416, 23.5384,)

2. (0.276348, 0.46651)

3. the sample size = 170.73 approximately 171

4. sample size = 334.07 approximately 334

5. sample of 334 should be taken by manager

Step-by-step explanation:

mean = bar x = 21.34 dollars

size of sample n = 70

standard deviation of sample = 9.22

we use t distribution as the population standard deviation is not known.

95% Confidence interval

1-α = 0.95

α = 0.05

degree of freedom = 70-1 = 69

α/2 = 0.025

using the t distribution tsble,

= 1.9949

confidence interval = 21.34+-1.9949*[\frac{9.22}{sqrt(70)}] \\

= 21.34 +- (1.9949*1.10200)

= 21.34 + 2.1984, 21.34 - 2.1984

= (23.5384, 19.1416)

the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384

2. sixe of those who only have a cat

p = 26/70 = 0.371429

at 90 % confidence interval,

1-α = 0.90

α = 0.10

we use the z table here

z(0.10/2) = Z(0.05)

= 1.645

0.371429+-1.645\sqrt} \frac{0.371429(1-0.371429)}{70}

= 0.371429 +-( 1.645 x 0.0578)

= 0.371429 + 0.095081, 0.371429 - 0.095081

= (0.276348, 0.46651)

3. sd = 10$

margin of erro,r e = 1.50$

α = 0.05

using z table

α/2 = Z0.025

= 1.96

sample size = 1.96² * 10² / 1.50²

= 3.8416 * 100/ 2.25

= 170.73

the sample size is approximately 171

d. we have 0.5 as sample proportion now

margin of error = 0.045

α = 0.10

Zα/2 = 0.05

= 1.645

sample size = 1.645²x0.5(1-0.5) / 0.045²

= 0.676506/0.002025

= 334. 07

sample size = 334

5.  sample of 334 should be taken by manager

3 0
3 years ago
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