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3 years ago

9

A bottle contains 2 green gum balls, 2 red gum balls and 1 blue gum ball. What is the probability of choosing a random gum ball

from the bottle that is NOT green? *

1 answer:

Ulleksa [173]3 years ago

6 0

Answer:

30%

Step-by-step explanation:

You might be interested in

What is the written form for (8,8)

Sav [38]

Answer:

81 900 000 000 000 is 8.19 × 10¹³

Step-by-step explanation:

Write the first number 8.

Add a decimal point after it: 8.

Now count the number of digits after 8. There are 13 digits.

So, in standard form: 81 900 000 000 000 is 8.19 × 10¹³

5 0

3 years ago

a larger gear wheel has 40 teeth and a small gear wheel has 25 teeth. if the larger one makes 100 revolutions in a minute, how m

nevsk [136]

Lets calculate gear ration T2/T1
so the gear ratio will be 1.6 it simply means dat if u rotate large gears one full revolution it tends to turn another short wheel to 1.6 revs, so in ur qtion the large whell revs 100 den u must multiply wid 1.6*100= 160 revs

5 0

3 years ago

Rachel and her friends eat 5/4 pizzas. How can you write the amount of pizza they ate as a mixed number?

andrew-mc [135]

Divide 5 by 4 and the left over Is the top of the fraction then your divisor is on the bottom so it would be 1 and 4/5.

6 0

3 years ago

Read 2 more answers

NO LINKS OR FILES. Explain how to evaluate 5p + (-6) when p = -4. Please give an explanation.

alexdok [17]

Answer:

-26

Step-by-step explanation:

5p + (-6)

Let p=-4

5(-4) + (-6)

Multiply

-20 +-6

Add

-26

8 0

3 years ago

Read 2 more answers

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$

which is about 651.918, so that the area is approximately $651.918\pi\approx\boxed{2048}$ .

Compare this to actual value of the integral, which is closer to 1967.

4 0

3 years ago