What is bigger 10 5/11 or 10. 5624

Hey guys, I need help with this word problem. I don't just want the answer. I would like the steps please!

ASHA 777 [7]

So, you take the equation for y given, y=0.28x+5.92. Then you substitute the values for x in, like so:
y=0.28(2)+5.92=6.42
y=0.28(5)+5.92=7.17
y=0.28(8)+5.92=7.92
So There's your table. Next:
7.88=0.28x+5.92
x=7, so the year when admission was approximately $7.88 is year 7.
Next,
10.04=0.28x+5.92
x=<span>14.7142857143, which rounds to year 15.
I hope this helps!</span>

7 0

2 years ago

Use the equation of the circle centered at the origin and

Serggg [28]

Answer:

Step-by-step explanation:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

If we are given x and y as 0 and 5 respectively, and we are also told that the center is (0, 0), our h and k are both 0. Filling in x, y, h, and k we can find the radius. So let's do that:

$(0-0)^2+(5-0)^2=r^2$ and

$0^2+5^2=r^2$ so

$r^2=25$ Our circle's equation is

$x^2+y^2=25$

Since the point in question, (4, 4), lies in the first quadrant, we will concentrate on that quadrant only. To fall within the circle, we can set up an inequality and test the point (4, 4). If it lies ON the circle then the equality would be true. Let's try that first:

$4^2+4^2=25$

Obviously, 16 + 16 does not equal 25, so that point (4, 4) does not lie ON the circle. In fact from that statement alone, we can determine that the point lies outside the circle because

$4^2+4^2>25$

If the inequality < were true then the point would lie inside.

5 0

3 years ago

How do you solve this question? Any help appreciated!

fiasKO [112]

$\bf \displaystyle \int\limits_{1}^{e}\cfrac{1}{t}\cdot dt\\\\
-------------------------------\\\\
\textit{doing substitution}\\\\
u=\cfrac{1}{t}\implies u=t^{-1}\implies \cfrac{du}{dt}=-t^{-2}\implies \cfrac{du}{dt}=-\cfrac{1}{t^2}\\\\\\ -t^2du=dt\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -t^2du\impliedby \textit{now, let's do some substitution on the "t"}\\\\
-------------------------------\\\\$

$\bf u=\cfrac{1}{t}\implies t=\cfrac{1}{u}\implies t^2=\cfrac{1^2}{u^2}\implies t^2=\cfrac{1}{u^2}\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -\cfrac{1}{u^2}\cdot du\implies -1\int\limits_{1}^{e}\cfrac{1}{u}\cdot du\implies \left. -ln|u| \cfrac{}{}\right]_1^e$

$\bf \left. -ln\left( \frac{1}{t} \right) \cfrac{}{}\right]_1^e\implies 
\left[ -ln\left( \frac{1}{e} \right) \right]-\left[ -ln\left( \frac{1}{1} \right) \right]\implies 
\left[ -ln\left( e^{-1}\right) \right]-\left[ -ln\left( 1\right) \right]
\\\\\\\
[-(-1)]-[-(0)]\implies 1-0\implies 1$

8 0

3 years ago

A bag contains 6 red balls and 5 black balls . Two balls are picked , one after the other . Find the probability that :

Kaylis [27]

Answer:

<u>There are:</u>

6 red balls - R
5 black balls - B
Total number = 11 balls

<h3>A. Without replacement</h3>

i. <u>Two blacks </u>

P(BB) = 5/11*4/10 = 2/11

ii. <u>The first is black</u>

P(B) = 5/11

or, alternatively

P(BR or BB) = 5/11*6/10 + 2/11 = 3/11 + 2/11 = 5/11

iii. <u>Both are of same colour</u>

P(BB or RR) = 2/11 + 6/11*5/10 = 2/11 + 3/11 = 5/11

<h3>B. With replacement</h3>

i. <u>Two blacks </u>

P(BB) = 5/11*5/11 = 25/121

ii. <u>The first is black</u>

P(B) = 5/11

or alternatively

P(BR or BB) = 5/11*6/11 + 25/121 = 30/121 + 25/121 = 55/121 = 5/11

iii. <u>Both are of same colour</u>

P(BB or RR) = 5/11*5/11 + 6/11*6/11 = 25/121 + 36/121 = 61/121

7 0

2 years ago

A scooter travels 30 feet in 2 seconds at a constant speed.

Sever21 [200]

Answer:

The scooter travels 15 feet in 1 second

The scooter travels 45 feet in 3 seconds

The scooter travels 60 feet in 4 seconds

The scooter travels 75 feet in 5 seconds

Step-by-step explanation:

First, let's find the number of feet the scooter travels in 1 second

(30 : 2) ÷ 2 = 15 : 1

Now that we have the unit rate, we can easily find the answers to the other questions.

(15 : 1) x 3 = 45 : 3

(15 : 1) x 4 = 60 : 4

(15 : 1) x 5 = 75 : 5

I Hope That This Helps! :)

8 0

2 years ago