Use pictures to explain how to use the sum 13 to show how to add in any order

Daniel works no more than 16 hours a week while attending college. He wants to spend at least 2 hours but no more than 10 hours

Vinvika [58]

Answer:

The maximum earning is $122

Step-by-step explanation:

The computation of the Daniel maximum earnings is as follows;

Since there is total of 16 hours

so 10 hours would be for parking cars and the remainig hours i.e.

= 16 hours - 10 hours

= 6 hours

This above hours for working at store

So,

= 10 hours × $8 + 6 hours × $7

= $80 + $42

= $122

Hence, the maximum earning is $122

6 0

3 years ago

Need help please show ur work

boyakko [2]

Use distributive property
5(4x + 3) - 2x
20x + 15 - 2x
Simplify by liked terms
18x + 15
The solution is A

3 0

3 years ago

A sports store has 468 golf balls. They will be put into boxes that hold 18 balls each. What is the minimum number of boxes need

Alona [7]

468/18 = 26

Therefore they need at least 26 boxes to hold all the golf balls.

Hope this helps :)

7 0

3 years ago

Read 2 more answers

The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

3 0

3 years ago

The weather forecast for the weekend is a 34% chance of rain for Saturday and a 32% chance of rain for Sunday. If we assume that

kaheart [24]

Answer:

$P( A \cap B) = P(A) *P(B) = 0.34*0.32 = 0.1088$

And then replacing in the total probability formula we got:

$P(A \cup B) = 0.34+0.32 - 0.1088 = 0.5512$

And rounded we got $P(A \cup B ) = 0.551$

That represent the probability that it rains over the weekend (either Saturday or Sunday)

Step-by-step explanation:

We can define the following notaton for the events:

A = It rains over the Saturday

B = It rains over the Sunday

We have the probabilities for these two events given:

$P(A) = 0.34 , P(B) = 0.32$

And we are interested on the probability that it rains over the weekend (either Saturday or Sunday), so we want to find this probability:

$P(A \cup B)$

And for this case we can use the total probability rule given by:

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

And since we are assuming the events independent we can find the probability of intersection like this:

$P( A \cap B) = P(A) *P(B) = 0.34*0.32 = 0.1088$

And then replacing in the total probability formula we got:

$P(A \cup B) = 0.34+0.32 - 0.1088 = 0.5512$

And rounded we got $P(A \cup B ) = 0.551$

That represent the probability that it rains over the weekend (either Saturday or Sunday)

6 0

3 years ago