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mafiozo [28]
3 years ago
12

Tony will spend at most $27 on gifts. So far, he has spent $14. What are the possible additional amounts he will spend?

Mathematics
2 answers:
AleksandrR [38]3 years ago
7 0

Answer:

27-14=c

Step-by-step explanation:

Also can you give brainliest to all of your questions bc it really hard to get brainliest and it be really nice you can choose who i dont really care but just please do it it helps a lot

Julli [10]3 years ago
3 0

Answer:

Inequaltiy: 27 ≤ 14+c

Final answer: 13 ≤ c

Step-by-step explanation:

Well tony spent at most 27 dollars.

He spent 14 already and he can‘t get past 27 dollars but he can spend exactly 27 or even less.

So 27 ≤ 14+c

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Write the slope-intercept form of the line that passes through (-10,17) and (5,-4)
luda_lava [24]

Answer:

y = 7/5 x + 31

Step-by-step explanation:

To find the slope, we need to use the formula

m = (y2-y1)/(x2-x1)

m =(-4-17)/(5- -10)

   = -21/(5+10)

  = 21/15

Divide the top and bottom by 3.

  = 7/5

The slope is 7/5.


Using the point slope form of the equation,

y-y1 = m(x-x1)

y-17 = 7/5 (x--10)

y-17 = 7/5(x+10)

Distribute the 7/5 ths.

y-17 = 7/5 x + 7/5*10

y-17 = 7/5 x +14

Add 17 to each side

y = 7/5 x + 14 + 17

y = 7/5 x + 31


This is in slope intercept form, with the slope  being 7/5 and the y intercept of 31

3 0
3 years ago
A box measures 20 inches wide by 14 inches deep. The height of the box is two feet. What is the volume of the box? Remember two
Jobisdone [24]

Answer:

6,720 inches or 560 feet

Step-by-step explanation:

20 x 14 x 24 = said number above.

Remember volume is just length times width times height.

8 0
3 years ago
Consider the following hypothesis test: H 0: 20 H a: < 20 A sample of 60 provided a sample mean of 19.5. The population stand
viva [34]

Answer:

We reject the null hypothesis and accept the alternate hypothesis. Thus, it be concluded that the population mean is less than 20.          

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 60

Sample mean, \bar{x} = 19.5

Sample size, n = 60

Alpha, α = 0.05

Population standard deviation, σ = 1.8

First, we design the null and the alternate hypothesis

H_{0}: \mu = 20\\H_A: \mu < 20

We use One-tailed z test to perform this hypothesis.

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Putting all the values, we have

z_{stat} = \displaystyle\frac{19.5 - 20}{\frac{1.8}{\sqrt{60}} } = -2.151

Now, z_{critical} \text{ at 0.05 level of significance } = -1.64

Since,  

z_{stat} < z_{critical}

We reject the null hypothesis and accept the alternate hypothesis. Thus, it be concluded that the population mean is less than 20.

5 0
4 years ago
Tonya is looking at a graph that shows a line drawn between two points with a slope of -5. One of the points is smudged and she
Molodets [167]

Answer:

5

Step-by-step explanation:

y=3x-5 if you graph it it will go through -5 and end up at 5,10

5 0
3 years ago
Your friend asks if you would like to play a game of chance that uses a deck of cards and costs $1 to play. They say that if you
gtnhenbr [62]

Answer:

Expected value = 40/26 = 1.54 approximately

The player expects to win on average about $1.54 per game.

The positive expected value means it's a good idea to play the game.

============================================================

Further Explanation:

Let's label the three scenarios like so

  • scenario A: selecting a black card
  • scenario B: selecting a red card that is less than 5
  • scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

  • 1/2 = 13/26
  • 2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}

Then we add up the results of that fourth column to compute the expected value.

(1/2) + (18/13) + (-9/26)

13/26 + 36/26 - 9/26

(13+36-9)/26

40/26

1.538 approximately

This value rounds to 1.54

The expected value for the player is 1.54 which means they expect to win, on average, about $1.54 per game.

Therefore, this game is tilted in favor of the player and it's a good decision to play the game.

If the expected value was negative, then the player would lose money on average and the game wouldn't be a good idea to play (though the card dealer would be happy).

Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

7 0
2 years ago
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