In circle F, arc AD = 80 and arc BC = 120.

(1,46) (2,40) (3,39) (4,35) (5,30) (6,27) equation line of best fit

siniylev [52]

Answer:

ertyghujikol

Step-by-step explanation:

3456t7ujn

8 0

3 years ago

Your sock drawer has two white socks, four brown socks, and two black socks. You randomly pick a sock and put it on your left fo

zmey [24]

Answer:

The probability of this occurring is 1/7

Step-by-step explanation:

Okay, here is a probability question.

From the question, we can identify the following;

Total number of socks = 2 + 4 + 2 = 8 socks

Since we are not given the order in which the socks was worn, we can assume any order.

Let’s say the right leg socks was worn first.

From the question, the socks on the right leg is a brown one.

So now let’s calculate the probability of selecting a brown socks out of the mix

That would be ;

number of brown socks/ total number of socks = 4/8 = 1/2

Now, for the left foot, we are having white sock here.

Kindly recall that since we already have a sock on the right foot, we have 7 socks left to pick from.

Now we need the probability of picking a white sock from the total 7. That would be ; 2/7

So the probability of both action occurring is simply the product of both probabilities and that is ;

1/2 * 2/7 = 1/7

6 0

3 years ago

What would x be equal to?

Vilka [71]

Answer:

x = $\frac{C-By}{A}$

Step-by-step explanation:

Given

Ax + By = C ( subtract By from both sides )

Ax = C - By ( divide both sides by A )

x = $\frac{C-By}{A}$

4 0

3 years ago

Help againnn. Yes, I’m kinda slow

Leni [432]

So, the right option is 1. so if my ans was helpful u can follow me.

5 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17$

8 0

3 years ago