Mathematics: If c= 23 and B=69° find a. round to the nearest tenth

If c= 23 and B=69° find a. round to the nearest, №8491242, 27.04.2023 17:52

Physics

1. An arrow is shot with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. What is the maximum height it will reach? a. 23 m b. 46 m c. 69 m d. 92 m 2. A projectile is launched horizontally from a cliff with an initial speed of 40 m/s. The cliff is 125 m high and the projectile travels a horizontal distance of 200 m from the bottom of the cliff. What is the speed of the projectile right before it hits the ground? a. 48 m/s b. 52 m/s c. 56 m/s d. 60 m/s e. 64 m/s

Step-by-step answer

P Answered by PhD

46m

Explanation:

Given the following :

Initial Velocity (u) = 60

Angle(θ) = 30°

Maximum height it will reach?

The maximum height if a projection is calculated using the formula :

H = u²sin²θ / 2g

Where H = maximum height, g = acceleration due to gravity = 9.8m/s²

H = (60²sin × 0. 5²) / 2(9.8)

H = 900 / 19.6

H = 45.918m

H = 46m

Physics

1. An arrow is shot with an initial velocity of 60.0 m/s at an angle of 30.0° above the horizontal. What is the maximum height it will reach? a. 23 m b. 46 m c. 69 m d. 92 m 2. A projectile is launched horizontally from a cliff with an initial speed of 40 m/s. The cliff is 125 m high and the projectile travels a horizontal distance of 200 m from the bottom of the cliff. What is the speed of the projectile right before it hits the ground? a. 48 m/s b. 52 m/s c. 56 m/s d. 60 m/s e. 64 m/s

Step-by-step answer

P Answered by PhD

46m

Explanation:

Given the following :

Initial Velocity (u) = 60

Angle(θ) = 30°

Maximum height it will reach?

The maximum height if a projection is calculated using the formula :

H = u²sin²θ / 2g

Where H = maximum height, g = acceleration due to gravity = 9.8m/s²

H = (60²sin × 0. 5²) / 2(9.8)

H = 900 / 19.6

H = 45.918m

H = 46m

Mathematics

URGENT, NEED ANSWERS ASAP 1) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p n = 110, x=55, 88% confidence a. 0.426 p 0.574 b. 0.422 p 0.578 c. 0.425 p 0.575 d. 0.421 p 0.579 2) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p A study involves 669 randomly selected deaths, with 31 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.

Step-by-step answer

P Answered by PhD

4

1) a. CI = 0.426 < p < 0.574

2) c. CI = 2.74% < p < 6.524%

3) a. CI = 20 < μ < 22

4) c. CI = 81.11 < μ < 85.09

5) b. CI = 76.21 < μ < 89.79

6) d. $453.59 < μ < $874.69

7) d. CI = $559 < σ < $1953.3

8) c. CI = $3.96 < σ < $6.72

9) c. 6.22 ft < σ < 12.59 ft

10) d. 0.19 oz < σ < 0.72 oz

Step-by-step explanation:

The Confidence Interval, CI are given as follows

For the population proportion

$CI=\hat{p}\pm z\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

1) $\hat p$ = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2) $\hat p$ = 31/669 = 0.5

z at 98% = ±2.326

c. CI = 2.74% < p < 6.524%

3) Here we have unknown population standard deviation, so we find the t interval

$\bar x$ = 21

n = 130

s = 3.0

Confidence level = 98%

$CI=\bar{x}\pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

$t_{\alpha /2}$ = ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

$t_{\alpha /2}$ = ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

$\bar x$ = 83

s = 13.5

n = 30

Confidence level = 99%

$t_{\alpha /2}$ = ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

$\bar x$ = $664.14

s = $297.29

n = 14

Confidence level = 98%

$t_{\alpha /2}$ = ±2.65 and

CI = $453.56 < μ < $874.72

d. $453.59 < μ < $874.69

7) The confidence interval for a population standard deviation is given by the following relation;

$\sqrt{\frac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\frac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

Here

n = 9

x = $3959

s = $886

Therefore we have

d. CI = $559 < σ < $1953.3

8) Here we have;

n = 41

x = $108

s = $5

Therefore;

c. CI = $3.96 < σ < $6.72

9) Here we have;

n = 29

x = 65 ft

s = 8.5 ft

Therefore we have

CI = 6.3 ft < σ < 12.73 ft

c. 6.22 ft < σ < 12.59 ft

10) Here we have

n = 8

s = 0.301188123

d. 0.19 oz < σ < 0.72 oz.

Mathematics

URGENT, NEED ANSWERS ASAP 1) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p n = 110, x=55, 88% confidence a. 0.426 p 0.574 b. 0.422 p 0.578 c. 0.425 p 0.575 d. 0.421 p 0.579 2) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p A study involves 669 randomly selected deaths, with 31 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.

Step-by-step answer

P Answered by PhD

4

1) a. CI = 0.426 < p < 0.574

2) c. CI = 2.74% < p < 6.524%

3) a. CI = 20 < μ < 22

4) c. CI = 81.11 < μ < 85.09

5) b. CI = 76.21 < μ < 89.79

6) d. $453.59 < μ < $874.69

7) d. CI = $559 < σ < $1953.3

8) c. CI = $3.96 < σ < $6.72

9) c. 6.22 ft < σ < 12.59 ft

10) d. 0.19 oz < σ < 0.72 oz

Step-by-step explanation:

The Confidence Interval, CI are given as follows

For the population proportion

$CI=\hat{p}\pm z\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

1) $\hat p$ = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2) $\hat p$ = 31/669 = 0.5

z at 98% = ±2.326

c. CI = 2.74% < p < 6.524%

3) Here we have unknown population standard deviation, so we find the t interval

$\bar x$ = 21

n = 130

s = 3.0

Confidence level = 98%

$CI=\bar{x}\pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

$t_{\alpha /2}$ = ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

$t_{\alpha /2}$ = ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

$\bar x$ = 83

s = 13.5

n = 30

Confidence level = 99%

$t_{\alpha /2}$ = ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

$\bar x$ = $664.14

s = $297.29

n = 14

Confidence level = 98%

$t_{\alpha /2}$ = ±2.65 and

CI = $453.56 < μ < $874.72

d. $453.59 < μ < $874.69

7) The confidence interval for a population standard deviation is given by the following relation;

$\sqrt{\frac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\frac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

Here

n = 9

x = $3959

s = $886

Therefore we have

d. CI = $559 < σ < $1953.3

8) Here we have;

n = 41

x = $108

s = $5

Therefore;

c. CI = $3.96 < σ < $6.72

9) Here we have;

n = 29

x = 65 ft

s = 8.5 ft

Therefore we have

CI = 6.3 ft < σ < 12.73 ft

c. 6.22 ft < σ < 12.59 ft

10) Here we have

n = 8

s = 0.301188123

d. 0.19 oz < σ < 0.72 oz.

Mathematics

1) find the mean for the given sample data. unless indicated otherwise, round your answer to one more decimal place than is present in the original data values. the amount of time (in hours) that sam studied for an exam on each of the last five days is given below. 1.7 7.7 8.3 1.6 5.1 find the mean study time a) 4.96 hr b) 5.45 hr c) 4.88 hr d) 24.40 hr 2) in order to avoid staying after work, a quality control analyst inspects the first 100 items produced that day. which sampling method did she use? a) cluster b) systematic c) stratified d) convenience

Step-by-step answer

P Answered by Master

4

C) 4.88 hr; D) convenience; B) P(A and B); A) 72; B) mean = 27.5, range = 20; no function shown to answer the question; A) 6.58, 0.42; D) average class size, cost; D) 7.8

Step-by-step explanation:

#1) To find the mean, we add all of the numbers and divide by the number of data values:

(1.7+7.7+8.3+1.6+5.1)/5 = 24.4/5 = 1.88

#2) This is a convenience sample because it was easy for the analyst to do; she did not use random sampling or any sort of groups, and she did not choose every kth item.

#3) The probability of two events is P(A and B) by definition.

#4) To find the total number of choices, we multiply the number of flavors by the number of toppings:

6(12) = 72

#5) To find the mean, add all of the data values up and divide by the number of data values:

(26+19+23+39+31+34+23+25)/8 = 220/8 = 27.5

To find the range, subtract the highest and lowest values:

39-19 = 20

#6) There is no function given to answer the question.

#7) Using the quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=\frac{--28\pm \sqrt{(-28)^2-4(4)(11)}}{2(4)}\\\\=\frac{28\pm \sqrt{784-176}}{8}\\\\=\frac{28\pm \sqrt{608}}{8}\\\\=\frac{28\pm 24.65766}{8}\\\\=\frac{28+24.65766}{8},\frac{28-24.65766}{8}\\\\=\frac{52.65766}{8},\frac{3.34234}{8}\\\\=6.58,0.42$

#8) The independent quantity is the one that causes the dependent to change. In this case, the average class size causes the cost of books to change; this means the average class size is independent and the cost of books is dependent.

#9) Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\=\sqrt{(8-2)^2+(3-8)^2}\\\\=\sqrt{6^2+(-5)^2}\\\\=\sqrt{36+25}=\sqrt{61}=7.8$

Mathematics

1) find the mean for the given sample data. unless indicated otherwise, round your answer to one more decimal place than is present in the original data values. the amount of time (in hours) that sam studied for an exam on each of the last five days is given below. 1.7 7.7 8.3 1.6 5.1 find the mean study time a) 4.96 hr b) 5.45 hr c) 4.88 hr d) 24.40 hr 2) in order to avoid staying after work, a quality control analyst inspects the first 100 items produced that day. which sampling method did she use? a) cluster b) systematic c) stratified d) convenience

Step-by-step answer

P Answered by Specialist

4

C) 4.88 hr; D) convenience; B) P(A and B); A) 72; B) mean = 27.5, range = 20; no function shown to answer the question; A) 6.58, 0.42; D) average class size, cost; D) 7.8

Step-by-step explanation:

#1) To find the mean, we add all of the numbers and divide by the number of data values:

(1.7+7.7+8.3+1.6+5.1)/5 = 24.4/5 = 1.88

#2) This is a convenience sample because it was easy for the analyst to do; she did not use random sampling or any sort of groups, and she did not choose every kth item.

#3) The probability of two events is P(A and B) by definition.

#4) To find the total number of choices, we multiply the number of flavors by the number of toppings:

6(12) = 72

#5) To find the mean, add all of the data values up and divide by the number of data values:

(26+19+23+39+31+34+23+25)/8 = 220/8 = 27.5

To find the range, subtract the highest and lowest values:

39-19 = 20

#6) There is no function given to answer the question.

#7) Using the quadratic formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=\frac{--28\pm \sqrt{(-28)^2-4(4)(11)}}{2(4)}\\\\=\frac{28\pm \sqrt{784-176}}{8}\\\\=\frac{28\pm \sqrt{608}}{8}\\\\=\frac{28\pm 24.65766}{8}\\\\=\frac{28+24.65766}{8},\frac{28-24.65766}{8}\\\\=\frac{52.65766}{8},\frac{3.34234}{8}\\\\=6.58,0.42$

#8) The independent quantity is the one that causes the dependent to change. In this case, the average class size causes the cost of books to change; this means the average class size is independent and the cost of books is dependent.

#9) Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\=\sqrt{(8-2)^2+(3-8)^2}\\\\=\sqrt{6^2+(-5)^2}\\\\=\sqrt{36+25}=\sqrt{61}=7.8$

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Physics

14. a block rests on a frictionless table on earth. after a 20-n horizontal force is applied to the block, it accelerates at 3.9 m/s2. then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. a horizontal force of 10 n is then applied to the block. what is the acceleration? a. 1.8 m/s2 b. 2.5 m/s2 c. 2.1 m/s2 d. 2.3 m/s2 e. 2.0 m/s2 15. workers are trying to free an suv stuck in the mud. to move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure.

Step-by-step answer

P Answered by PhD

5

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg , where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

Chemistry

20 match the alien matters with the correct on on earth: periodic table group 1 group 2 group 13 group 14 group 15 group 16 group 17 group 18 1 h/gs 1.008 2 he/ 4.003 3 li/ 7.941 4 be/ 9.012 5 b/ 10.811 6 c/ 12.010 7 n/at 14.006 8 o 15.999 9 f/ 18.998 10 ne/ 20.179 11 na/so 22.989 12 mg/ 24.305 13 al/lm 26.981 14 si/ 28.085 15 p/ 30.973 16 s/ 32.065 17 cl/ms 35.453 18 ar/ 39.948 19 k/ 39.098 20 ca/ 40.007 skip 21 - 30 31 ga/ 69.723 32 ge 72.64 33 as/ 74.921 34 se/ 78.96 35 br/ 79.904 36 kr/ 83.8 metals metalloids nonmetals **galactic element clues**

Step-by-step answer

P Answered by Specialist

34

H=Gs, Li=Me, Na=So, K=Bn, Be=Am, Mg=A, Ca=Bo, B=Sc, Al=Lm, Ga=Sk, C=Or, Si=Co, Ge=Ik, N=At, P=Fw, As=Ki, O=Bt, S=Re, Se=Uv, F=T, Cl=Ms, Br=Lq, He=Fl, Ne=Lg, Ar=Nb, and Kr=Sp.

I may skipped one by mistake, let me know if this answers everything or if you need any other help with it.

Chemistry

20 match the alien matters with the correct on on earth: periodic table group 1 group 2 group 13 group 14 group 15 group 16 group 17 group 18 1 h/gs 1.008 2 he/ 4.003 3 li/ 7.941 4 be/ 9.012 5 b/ 10.811 6 c/ 12.010 7 n/at 14.006 8 o 15.999 9 f/ 18.998 10 ne/ 20.179 11 na/so 22.989 12 mg/ 24.305 13 al/lm 26.981 14 si/ 28.085 15 p/ 30.973 16 s/ 32.065 17 cl/ms 35.453 18 ar/ 39.948 19 k/ 39.098 20 ca/ 40.007 skip 21 - 30 31 ga/ 69.723 32 ge 72.64 33 as/ 74.921 34 se/ 78.96 35 br/ 79.904 36 kr/ 83.8 metals metalloids nonmetals **galactic element clues**

Step-by-step answer

P Answered by Master

34

H=Gs, Li=Me, Na=So, K=Bn, Be=Am, Mg=A, Ca=Bo, B=Sc, Al=Lm, Ga=Sk, C=Or, Si=Co, Ge=Ik, N=At, P=Fw, As=Ki, O=Bt, S=Re, Se=Uv, F=T, Cl=Ms, Br=Lq, He=Fl, Ne=Lg, Ar=Nb, and Kr=Sp.

I may skipped one by mistake, let me know if this answers everything or if you need any other help with it.

If c= 23 and B=69° find a. round to the nearest tenth

Faq

Try asking the Studen AI a question.