All categories

3 years ago

Which is an advantage of using a checking account?

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

7 0

You don’t have to worry about identity thief

bulgar [2K]3 years ago

6 0

B i believe, i hope this helped!

You might be interested in

What's y=950-23x as an system of linear equations

sdas [7]

Your answer is below (for y)
y=−23x+950

Your answer for x is

x=−y/23+950/23x

6 0

3 years ago

emma lee is shopping for supplies for her science project. The regular price of the supplies at a hardware store is $20. The sto

zmey [24]

Answer:

First, you add 30 and 20, and you get 50. Then, since 30 is your highest priced item, you find 0.30 of 30, which is 9. After that you would subtract 9 from 30, which gives you 21. Therefore, we know the price is 41

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

(x + 3)(x -4) simplify

irga5000 [103]

Answer:

your answer is x^2 - x - 12

3 0

3 years ago

Read 2 more answers

You insurance policy on your car costs $600 a year. If you total your vehicle, you insurance company will pay $25,000 towards a

vladimir2022 [97]

Answer:

Your annual expected value is -$550.

Step-by-step explanation:

0.002 probability that you will total your car.

If you total your car, you pay the insurance of $600, but you get the $25,000 insurance. So in total, the net value will be $24,400.

1-0.002 = 0.998 probability that you will not total your car.

In this case, you lose $600.

What is your annual expected value?

Multiply each net value by it's probability.

E = 0.002*24400 - 0.998*600 = -550

Your annual expected value is -$550.

4 0

3 years ago

Please helpppp it has to do with P(A U B)

Katyanochek1 [597]

$P(A)=0.5$ , $P(B)=0.6$ , $P(A\cup B)=0.8$

We find the probability of intersection using the inclusion/exclusion principle:

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.3$

By definition of conditional probability,

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{0.3}{0.6}=0.5$

For $A$ and $B$ to be independent, we must have

$P(A\cap B)=P(A)\cdot P(B)$

in which case we have $0.3=0.5\cdot0.6$ , which is true, so $A$ and $B$ are indeed independent.

Or, to establish independence another way, in terms of conditional probability, we must have

$P(A\mid B)\cdot P(B)=P(A)\cdot P(B)\implies P(A\mid B)=P(A)$

which is also true.

7 0

3 years ago