All categories

3 years ago

Question 5

Mathematics

1 answer:

Sveta_85 [38]3 years ago

7 0

Answer:

Step-by-step explanation:

Use the dot plot to make a list of the numbers it represents, sorted from smallest to largest.

0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 9

There are 23 numbers in this list. The median is the one in the middle, so there will be 11 numbers to the left of it and 11 to the right.

The median is 2

You might be interested in

Nate has $50 to spend at the grocery store. he fills his shopping cart with items totaling $46. At checkout he will have to pay

Akimi4 [234]

Answer:

yes he can

Step-by-step explanation:

he has to pay total is : 46+46x6%=48.76$

so he can pay that

8 0

3 years ago

Read 2 more answers

Please help use pic.

sertanlavr [38]

Let the reduced price be x

If he paid for 6 rounds, he paid 1 round at full price of $19 and 5 rounds at reduced price:
19 + 5x = 59
5x = 59 - 19
5x = 40
x = 40 ÷ 5
x = 8

The reduced price is at $8 per round.

4 0

3 years ago

$180 car 15% discount

tatiyna

Answer:

$153

Step-by-step explanation:

180 * 0.85 = 153

3 0

3 years ago

Read 2 more answers

Please help with this

Degger [83]

All of them are wrong its 91 but if you want click or circle 90

3 0

3 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

3 years ago