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2 years ago

Find the value of X and give a reason for the relation ship between the angles.

Mathematics

1 answer:

Ostrovityanka [42]2 years ago

8 0

Answer:

x = 55

Step-by-step explanation:

(x - 14) + 99 + 220 = 360

x - 14 + 99 + 220 = 360

x - 14 + 319 = 360

x + 305 = 360

x = 360 - 305

x = 55

-TheUnknownScientist

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Eight over twelve equals x over thirty solve for x

xenn [34]

Answer:

8x12x30=2880

Step-by-step explanation:

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3 years ago

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2 years ago

When a particular mass-spring system is started by stretching the spring 3 cm, one complete oscillation takes 4 seconds.

11Alexandr11 [23.1K]

Answer:

A. 5.16 s.

B. 5.66 s.

Step-by-step explanation:

For a simple harmonic motion,

T = 2pi (sqrt * (l/g))

Given:

L1 = 3 cm

T1 = 4 s

L2 = 5 cm

T2 = ?

4 = 2pi*sqrt(3/g)

g = 7.4

At, L2,

T2 = 2pi*sqrt(5/7.4)

= 5.16 s.

M1 = M1

M2 = 2*M1

For a simple harmonic motion,

T = 2pi (sqrt * (m/k))

4 = 2pi (sqrt * (M1/k))

M1/k = 0.405

Inputting the above values,

T2 = 2pi (sqrt * (2*M1/k))

= 2pi (sqrt * (2 * 0.405))

= 5.66 s.

4 0

2 years ago

Spell out the name of a planet with a 29-day year

Jet001 [13]

The answer would be mercury.

6 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago