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3 years ago

40 points!!

Mathematics

1 answer:

vredina [299]3 years ago

5 0

Answer:

The translation is right 3 down 2 which is reflection

Step-by-step explanation:

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canning operation is designed to fill cans with 14.5 oz of ingredients, on average, with a standard deviation of 0.1 oz. Nine ca

Pavlova-9 [17]

Answer:

SE = 0.025

Step-by-step explanation:

We are given;

Sample mean; x¯ = 14.52

Sample standard deviation; s = 0.075

Sample size; n = 9

Now,formula for standard error of sample mean is;

SE = s/√n

SE = 0.075/√9

SE = 0.075/3

SE = 0.025

3 0

3 years ago

Your friend attempted to factor an expression as shown. Find the error in your friends work. Then factor the expression correctl

loris [4]

Answer:

Third step is incorrect. The correct factored form is (x-1)(2x-5).

Step-by-step explanation:

The given expression is

$2x^2-7x+5$

We need to find the factored form of this expression.

Step 1: Given

$2x^2-7x+5$

Step 2: Splitting the middle term method, the middle term can be written as (-5x-2x).

$2x^2-5x-2x+5$

$(2x^2-5x)+(-2x+5)$

Step 3: Taking out common factors from each parenthesis.

$x(2x-5)-1(2x-5)$

Step 4: Taking out common factors.

$(x-1)(2x-5)$

Therefore, the third step is incorrect. The correct factored form is (x-1)(2x-5).

3 0

2 years ago

Solve for x.

statuscvo [17]

Answer:

The Answer is: x = 1.444 with repeating 4.

Step-by-step explanation:

212x - 34(2x + 5) = 38

212x - 68x - 170 = 38

212x - 68x = 38 + 170

144x = 208

x = 208/144 = 52/35 or 1.444

Proof:

212(1.444) - 34(2(1.444) + 5) = 38

306.128 - 34(7.888) = 38

306.128 - 268.192 = 37.936 or approximately 38

Hope this helps! Have an Awesome Day!! :-)

7 0

2 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

2 years ago

Six more than the quotient of a number and 2 equals 8

ikadub [295]

Let number=x

(X/2)+6=8

First subtract 6 on each side to isolate X on one side.

(X/2)=2

Now to fully isolate X by itself, just multiply 2 on each side.

X=4

7 0

3 years ago

Read 2 more answers