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3 years ago

Increase 100kg by 10%

Mathematics

1 answer:

KiRa [710]3 years ago

6 0

Answer:

110 kg

Step-by-step explanation

a simple way is to just do 100 plus 10= 110

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I will give out brainliest answer this

Archy [21]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Philip ran out of time while taking a multiple-choice test and plans to guess the last 4 questions. Each question has 5 possible

White raven [17]

Using the binomial distribution, it is found that there is a 0.4096 = 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

Considering that there are 4 questions, and each has 5 choices, the parameters are given as follows:

n = 4, p = 1/5 = 0.2.

The probability that he answers exactly 1 question correctly in the last 4 questions is P(X = 1), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{4,1}.(0.2)^{1}.(0.8)^{3} = 0.4096$

0.4096 = 40.96% probability that he answers exactly 1 question correctly in the last 4 questions.

More can be learned about the binomial distribution at brainly.com/question/24863377

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8 0

2 years ago

There are four white socks and six black socks in a drawer. What is the probability of selecting a white sock, not replacing it,

m_a_m_a [10]

Answer:

A: 1/3

Step-by-step explanation:

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8 0

3 years ago

Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|

Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0

3 years ago

8) through: (-3, -2), perp. to y = x – 1<br> A) y=-5x – 1 B) y=-4x – 5<br> C) y=-x – 5 D) y=-5x – 4

nexus9112 [7]

<u>Answer:</u>

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

<u>Solution:</u>

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1$

We know that, <em>product of slopes of perpendicular lines is -1. </em>

So, 1 $\times$ slope of perpendicular line = -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

8 0

3 years ago