Please it’s really important I get an answer fast

Sam had an appointment 90 miles from home at 2 pm. He drove to his appointment at average speed 60 mph and was 15 minutes late.

LUCKY_DIMON [66]

Answer:

He drove for 1 hour and 30 minutes, and started driving at 12:45pm

Step-by-step explanation:

Firstly, you deduct 60 and 90. That means that he has been driving one hour, then the remainder is 30, which means that at the speed of 60mph he traveled that much in half an hour. Then you add the 15 more minutes to the time of the meeting (don't let this confuse you, don't try to add the 15 more minutes to the time that he traveled, but to the end time). That means that he drove for a total of 1 hour and 30 minutes. Now you just move the clock back from 2:15 pm for 1 hour and 30 minutes. That means that he started driving at 12:45 pm

(this confused me so much and I don't know why, but I hope I helped)

:)

7 0

3 years ago

Four less than 3 times a number is the same as the number plus 10 - in an equation

ra1l [238]

Answer:

4-3x=x+10

Step-by-step explanation:

4- (four less)

3x (3 times a number)

= (is the same as)

x (the number)

+10 (plus ten)

Please rate, thanks, mark brainliest if this helps!

6 0

4 years ago

6x+4=2 Whats the value of x?

damaskus [11]

Step-by-step explanation:

-1/3 the answer is negative one third

5 0

3 years ago

Read 2 more answers

It is known that there only is 1% chance of getting a disease. a test is being devised to detect the disease. the probability th

Cerrena [4.2K]

Suppose $D$ is the event that a given patient has the disease, and $P$ is the event of a positive test result.

We're given that

$\mathbb P(D)=0.01$
$\mathbb P(P\mid D)=0.98$
$\mathbb P(P^C\mid D^C)=0.95$

where $A^C$ denotes the complement of an event $A$ .

a. We want to find $\mathbb P(P^C)$ . By the law of total probability, we have

$\mathbb P(P^C)=\mathbb P(P^C\cap D)+\mathbb P(P^C\cap D^C)$

That is, in order for $P^C$ to occur, it must be the case that either $D$ also occurs, or $D^C$ does. Then from the definition of conditional probability we expand this as

$\mathbb P(P^C)=\mathbb P(D)\mathbb P(P^C\mid D)+\mathbb P(D^C)\mathbb P(P^C\mid D^C)$

so we get

$\mathbb P(P^C)=0.01\cdot0.02+0.99\cdot0.95=0.9407$

b. We want to find $\mathbb P(D\mid P)$ . Now, we can use Bayes' rule, but if you're like me and you find the formula a bit harder to remember, we can easily derive it.

By the definition of conditional probability,

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D\cap P)}{\mathbb P(P)}$

We have the probabilities of $P$ / $P^C$ occurring given that $D$ / $D^C$ occurs, but not vice versa. However, we can expand the probability in the numerator to get a probability in terms of $P$ being conditioned on $D$ :

$\mathbb P(D\cap P)=\mathbb P(D)\mathbb P(P\mid D)$

Meanwhile, the law of total probability lets us rewrite the denominator as

$\mathbb P(P)=\mathbb P(P\cap D)+\mathbb P(P\cap D^C)$

or in terms of conditional probabilities,

$\mathbb P(P)=\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)$

so that

$\mathbb P(D\mid P)=\dfrac{\mathbb P(D)\mathbb P(P\mid D)}{\mathbb P(D)\mathbb P(P\mid D)+\mathbb P(D^C)\mathbb P(P\mid D^C)}$

which is exactly what Bayes' rule states. So we get

$\mathbb P(D\mid P)=\dfrac{0.01\cdot0.98}{0.01\cdot0.98+0.99\cdot0.05}\approx0.1653$

6 0

3 years ago

Divide these fractions. Cancel if necessary and simplify.can ya help me :)

jarptica [38.1K]

#2 9/10÷ 1/2=4/10 and number 3 is 2/4

3 0

3 years ago

Read 2 more answers