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s2008m [1.1K]
3 years ago
8

What is the percent of change from 25 to 18? Can you explain the steps for me pls and thanks!

Mathematics
2 answers:
levacccp [35]3 years ago
8 0

Answer:

18 is 72% of 25 which is also known as 28% less of 25

Step-by-step explanation:

18/25=.72

1-.72=.28

.28=28%

svp [43]3 years ago
4 0

Answer:

28%

Step-by-step explanation:

25 → 100%

1 → 100% / 25 = 4%

18 → 4% x 18 = 72%

percentage change = difference in percentage = 100% - 72% = 28%

hope that helps :)

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Lostsunrise [7]

Answer:

5

C. 2

6.

A. (4, 5)

7.B.broken

line.

8.A.above the line

9.. 2x + y < 4

10.D. (-2,-3)

4 0
2 years ago
Read 2 more answers
Please please help me. i’ll literally send you money. i need to pass please help
mihalych1998 [28]
<h3>Answer: Approximately 6.4 units</h3>

==================================================

Explanation:

The origin is the point (0,0)

Use the distance formula to find the distance from (0, 0) to (4, -5)

Let

(x_1,y_1) = (0,0)\\\\(x_2,y_2) = (4,-5)\\\\

be our two points. Plug those values into the distance formula below and use a calculator to compute

d = \sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0-(-5)\right)^2}\\\\d = \sqrt{\left(0-4\right)^2+\left(0+5\right)^2}\\\\d = \sqrt{\left(-4\right)^2+\left(5\right)^2}\\\\d = \sqrt{16+25}\\\\d = \sqrt{41} \ \text{ exact distance}\\\\d \approx 6.40312 \ \text{ approximate distance}\\\\d \approx 6.4\\\\

The distance between the two points (0,0) and (4,-5) is approximately 6.4 units.

6 0
3 years ago
Evaluate the following integrals: 1. Z x 4 ln x dx 2. Z arcsin y dy 3. Z e −θ cos(3θ) dθ 4. Z 1 0 x 3 √ 4 + x 2 dx 5. Z π/8 0 co
Zigmanuir [339]

Answer:

The integrals was calculated.

Step-by-step explanation:

We calculate integrals, and we get:

1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}

2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}

3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}

4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}

5)  \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=

=\frac{3π+8}{64}

6)  ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x

7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}

8)  ∫ tan^5 (x) sec(x)  dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x

6 0
3 years ago
What is the largest number that rounds to 560,000?
anastassius [24]

Answer:

559,999

Step-by-step explanation:


4 0
3 years ago
The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part
nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

z = 1.6

P-value from Z-Table:

P(x<48) = 0.9452

P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

4 0
3 years ago
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