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3 years ago

How many times can 26 go into 59

1 answer:

8 0

It can only go twice. 26 x 2 = 52, but 59 - 52 is 7, which is smaller than 26. So the final answer should be 2.
Hope this helped! :)

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I REALLY NEED HELP!!!!!!!

Reil [10]

Do you think the answer is C.147?

3 0

3 years ago

Graph the function: <br> f(x)=5/2x-6

MrMuchimi

Plug in any value for x and then you will get y. Use the coordinates to plot.
Example
x= 2
5/2(2)-6
5-6
-1
(2,-1)
Then continue to do the same.

5 0

3 years ago

A scuba diver was rising back to the surface after taking photos under water. At 2:50 P.M.,

djverab [1.8K]

Answer:

the answer is 11.6 feet

Step-by-step explanation:

you simply subtract the feet of the depth he was at at 2:55pm from his depth at 2:50pm

5 0

3 years ago

2. Is 4x – 3 = 19 true, false, or open? <br> A)true<br> B)false<br> C)open

ElenaW [278]

Answer: B. False

Step-by-step explanation: Because you do

4x - 3 = 19

+3 +3 =4x = 22

You divide 4x divide by 4 and you get x

Then you divide 22 by 4 and get 5.5

So the answer is X = 5.5

6 0

3 years ago

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2

Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is $\frac{1}{20000}$

Probability of getting second prize is $\frac{2}{20000}$

Probability of getting third prize is $\frac{20}{20000}$

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

$\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10$

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is $\frac{19977}{20000}$

The expected value is

$E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25$

$E=\frac{1000+600+200-4994.25}{20000}$

$E=\frac{-3194.25}{20000}$

$E=-0.159$

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0

3 years ago