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Keith_Richards [23]

3 years ago

13

A concession stand at a baseball game sells hot dogs and boxes of popcorn. Justin buys a total of 9 snacks. The number of popcor

n boxes he buys is three less than twice as many hot dogs. Which system of equations models this situation if he bought h hot dogs and p boxes of popcorn?

2 answers:

Stella [2.4K]3 years ago

6 0

Answer: did you ever get the answer? I have this same question on a test rn

Step-by-step explanation:

MAXImum [283]3 years ago

6 0

Answer:

}p=3−2h

][

}h=9−p

Step-by-step explanation:

i don't know the steps but i know the answer.

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Will give 30 points!<br> What is a expanded notation?

gavmur [86]

Answer:

It is the number written out to show each digit

Step-by-step explanation:

Let's say you have the number 459

you write out the hundredths place, which is 4 times 100

you write out the tenths place which is 5 times 10

and you write out the oneths place which is 9 times 1

the full thing written out is 459 = 100×4 + 5×10 + 9 × 1

4 0

3 years ago

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

3 0

3 years ago

Mr zambito buys a 3 1/2 - pound bag of dog food every 3 weeks mr zambito feeds his dog the same amount of dog food each day how

Murljashka [212]

3 1/2 pound = 7/2 pounds.
We know that the dog eats 7/2 (3.5) pounds every 21 days since there are 7 days in a week and Mr.Z only buys dog food every 3 weeks.

Set up a proportion. <u>3.5 lb</u> = <u> x </u><u>
</u> 21 days 365 days
<u>
</u>Cross multiply and divide. 3.5 lb x 365 days = 1277.5 / 21 days = 60.83 lb
<u>
</u>

3 0

3 years ago

Amelie found the solutions of the equation x2 − 1x = 42 to be 6 and −7. Explain why this answer is incorrect. Then, find the cor

cupoosta [38]

I dont know how to solve it but since the answer shows as 6 and -7, if you plug it in as x values, one tiem as 6 and second time as -7 it doesnt equal 42

7 0

3 years ago

Last July, 160 babies were born in a hospital in Maine; 3

Eduardwww [97]

Answer: C

Step-by-step explanation:

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