All categories

3 years ago

Which statement is true?

Mathematics

1 answer:

Sedbober [7]3 years ago

7 0

Answer:

This is the right one that's true

$\frac{2}{5 } = 0.4$

$8.6 = 0.086\%$

You might be interested in

3. What is the slope of the line that passes through the points E(5, 1) and F(2, -7)?

Veseljchak [2.6K]

It’s d because it negative and there is a negative 7

8 0

3 years ago

HELP PLEAAASEEE Ireland opened a bag or skittles and recorded the colors and their frequencies.

Anna11 [10]

The correct answer is 25%

<h3>What is Relative Frequency?</h3>

The number of times an event occurs divided by the total number of events occurring in a given data.

<h3>How to solve the problem?</h3>

This problem can be solved by following steps.
The data of color and frequency is given in table.
We need find the relative frequency for Purple

First calculate the the total number of frequency

18+20+10+16

= 64

The total number of frequency is 64

Hence the relative frequency of purple is 16/64

Therefore , relative frequency of purple is 1/4 = 0.25

Therefore 25% of purple candies are there

Learn more about Relative frequency here

brainly.com/question/3857836

#SPJ2

7 0

2 years ago

Read 2 more answers

before school began mrs. weeks bought a total of 86 balls us the information below to help you write a numerical expression

STALIN [3.7K]

For this case we have that the total of the balls is 86. We know there are 8 footballs then:

Basketball: Two more than twice the number of footballs are basketballs, that is:

$2 + 2 (8) = 2 + 16 = 18$

There are 18 Basketballs.

Baseballs: Four less than 5 times the number of footballs are baseballs, that is:

$5 (8) -4 = 40-4 = 36$

There are 36 Baseballs.

Softballs: Six more than half of baseballs are softballs. That is to say:

$6+ \frac {36} {2} = 6 + 18 = 24$

There are 24 Softballs

If we add we must get 86.

$8 + 18 + 36 + 24 = 86$

ANswer:

There are 8 Footballs

There are 18 Basketballs.

There are 36 Baseballs.

There are 24 Softballs

3 0

3 years ago

Help please......!! :(

Blababa [14]

6. X= -32 and 32
7. Y= -3
im not sure on the other questions and i LOVE doing math and im a sophmore in highschool so at least i could help get a couple of them

7 0

3 years ago

Read 2 more answers

For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso

PIT_PIT [208]

Answer:

$\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}$

$\textsf{Trapezium rule}: \quad \pi$

$\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}$

Step-by-step explanation:

<u>Midpoint rule</u>

$\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Trapezium rule</u>

$\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Simpson's rule</u>

$\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Given definite integral</u>:

$\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x$

Therefore:

a = 0
b = 2π

Calculate the subdivisions:

$\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi$

<u>Midpoint rule</u>

Sub-intervals are:

$\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]$

The midpoints of these sub-intervals are:

$\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi$

Therefore:

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}$

<u>Trapezium rule</u>

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}$

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}$

<u>Simpson's rule</u>

<u />

<u /> $\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}$

6 0

2 years ago