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2 years ago

Use the law of cosines round to the nearest tenths

Mathematics

1 answer:

PtichkaEL [24]2 years ago

3 0

Answer:

x = 11.5

Step-by-step explanation:

Formula

a^2 = b^2 + c^2 - 2*b*c * cos(A) Substitute givens in formula

Givens

b = 31
c = 26
A = 21o
a = x Let a = x to present the formula in it's most normal way

Solution

a^2 = 31^2 + 26^2 - 2 * 31 * 26 * cos(21) Find -2*31*26*cos(21)

a^2 = 961 + 676 -1504.93 Combine

a^2 = 1637 -1504.93 Combine again

a^2 = 132.07 Take the √ on both sides

√a^2 = √132.07

a = 11.5 Rounded.

Answer

x = 11.5

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Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true

IgorLugansk [536]

Answer:

(a) 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

Step-by-step explanation:

We are given that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation 0.75.

(a) Also, the average porosity for 20 specimens from the seam was 4.85.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.85

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 20

$\mu$ = true average porosity

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.85-1.96 \times {\frac{0.75}{\sqrt{20} } }$ , $4.85+1.96 \times {\frac{0.75}{\sqrt{20} } }$ ]

= [4.52 , 5.18]

Therefore, 95% confidence interval for the true average porosity of a certain seam is [4.52 , 5.18].

(b) Now, there is another seam based on 16 specimens with a sample average porosity of 4.56.

The pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average porosity = 4.56

$\sigma$ = population standard deviation = 0.75

n = sample of specimens = 16

$\mu$ = true average porosity

Here for constructing 98% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 98% confidence interval for the true mean, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98 {As the critical value of z at 1% level

of significance are -2.3263 & 2.3263}

P(-2.3263 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.3263$ ) = 0.98

P( $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.3263 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $4.56-2.3263 \times {\frac{0.75}{\sqrt{16} } }$ , $4.56+2.3263 \times {\frac{0.75}{\sqrt{16} } }$ ]

= [4.12 , 4.99]

Therefore, 98% confidence interval for the true average porosity of a another seam is [4.12 , 4.99].

7 0

3 years ago

Find L when P= 24 and W = 6. P = 2L+ 2W (perimeter of a rectangle) L =

Anon25 [30]

Answer:

Step-by-step explanation:

24 = 2L + 2(6), multiply...

24 = 2L + 12, subtract 12 to both sides...

12 = 2L, divide 2 to both sides to isolate the varibale L

6 = L

Therefroe, L = 6

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3 years ago

Factor by grouping 9m3-3m2p2-3mp+p3

Setler [38]

(3m^2-p)(3m-p^2)
You factor out the 3m^2 and p from the combined like terms and then group those into a factor and group the two identical factors together and (3m^2-p)(3m-p^2) is your answer.

3 0

4 years ago

Which statement describes the equation below?

Colt1911 [192]

Answer:

Abccd

Step-by-step explanation:

Hi low hin

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3 years ago

Both circle A and circle B have a central angle measuring 140°. The ratio of the radius of circle A to the radius of circle B is

beks73 [17]

Let
rA--------> radius of the circle A
rB-------> radius of the circle B
LA------> the length of the intercepted arc for circle A
LB------> the length of the intercepted arc for circle B

we have that
rA/rB=2/3--------> rB/rA=3/2
LA=(3/4)π

we know that
if Both circle A and circle B have a central angle , the ratio of the radius of circle A to the radius of circle B is equals to the ratio of the length of circle A to the length of circle B
rA/rB=LA/LB--------> LB=LA*rB/rA-----> [(3/4)π*3/2]----> 9/8π

the answer is
the length of the intercepted arc for circle B is 9/8π

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3 years ago