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2 years ago

Which polynomial is in standard form?

Mathematics

2 answers:

skelet666 [1.2K]2 years ago

7 0

Answer:

Step-by-step explanation:

Alchen [17]2 years ago

7 0

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The quotient of a number and -3 is at most 7

Grace [21]

Answer:

-21

Step-by-step explanation:

-3 * 7

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3 years ago

Read 2 more answers

Quien puede resolver solo el problema 59 , porfavor!!<br>si no van a responder bien , no respondan

RideAnS [48]

Answer:

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f(x)= 5/2x

a=8x/5

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2 years ago

This is page four what is the answer

Fittoniya [83]

Answer:

The answer is she lost 9 matches

Step-by-step explanation:

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2 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

A 4. 9-mm-wide swimming pool is filled to the top. The bottom of the pool becomes completely shaded in the afternoon when the su

cestrela7 [59]

The depth of the swimming pool that is filled to the top is; 4 m

<h3>Snell's Law</h3>

I have attached a schematic diagram showing this question.

The correct width of the pool is 4 meters. Thus; w = 4 m

Incident Angle; θ₁ = 20°

A right angle is 90° and so the angle θ₂ is calculated from;

θ₂ = 90° - θ₁

θ₂ = 90° - 20°

θ₂ = 70°

We can use snell's law formula to find θ₃.

Thus;

n₁sinθ₂ = n₂sinθ₃

where;

n₁ is refractive index of air = 1

n₂ is refractive index of water = 1.33

Thus;

1*sin 70 = 1.33*sin θ₃

sin θ₃ = (sin 70)/1.33

Solving this gives;

θ₃ = 44.95°

By usage of trigonometric ratios we can find the depth of the pool using;

w/d = tan θ₃

Thus;

d = w/(tan θ₃)

d = 4/(tan 44.95)

d ≈ 4 m

Read more about Snell's Law at; brainly.com/question/10112549

6 0

2 years ago