All categories

3 years ago

Please find the distance between the points

Mathematics

1 answer:

marusya05 [52]3 years ago

5 0

Answer:

they are 6 uints apart

Step-by-step explanation:

You might be interested in

Explain the differences (exponents)<br><br> -6^4 and (-6)^4

wel

Step-by-step explanation:

$- {6}^{4} = - 1296 \\ \\ (- {6})^{4} = 1296$

The product of $-6^4$ is negative whereas the product of $(-6) ^4 is positive.$

7 0

3 years ago

I need help plz help

ivanzaharov [21]

Answer:

B. 70%

Step-by-step explanation:

7/10 of the numbers are more than 5

7/10= 70%

5 0

3 years ago

Read 2 more answers

What’s the answer pls

larisa [96]

Amount of interest = 3/100 x 1200
= $36

Answer is B

3 0

3 years ago

Order these numbers for lest to greatest 1/2 ,1.1 ,5/3 ,2, 0 ,0.4, -2 ,5/8

Nat2105 [25]

The least number should always start off with negatives if there are any. In this case there is so we can start with -2 as our least amount. Next are the small fractions. If you're allowed a calculator, I would use that to find what the decimals will look like for each fraction:

$\frac{1}{2} =0.5\\ \\ \frac{5}{3} =1.6666666666666666666666666666667\\ \\ \frac{5}{8} =0.625$

After -2 the next least number is 0 and the next one is the fraction or decimal closest to the number 0 which is 0.4

The next number would be from the fractions we solved earlier. 0.5 and 0.625 would follow 0.4 and then 1.1 and 1.6667(solved for from fraction).

Here is the list:

$-2,0,0.4,\frac{1}{2} ,\frac{5}{8} ,1.1,\frac{5}{3}$

Hope this helps!

6 0

3 years ago

The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and co

julia-pushkina [17]

Answer:

P[X=3,Y=3] = 0.0416

Step-by-step explanation:

Solution:

- X is the RV denoting the no. of customers in line.

- Y is the sum of Customers C.

- Where no. of Customers C's to be summed is equal to the X value.

- Since both events are independent we have:

P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]

P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]

P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]

P[X=3]*P[C1=1,C2=1,C3=1] = P[X=3]*(P[C=1]^3)

- Thus, we have:

P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3

P[X=3,Y=3] = 0.0416

6 0

2 years ago