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Brilliant_brown [7]
3 years ago
9

Lawrence has a total of 15 five dollar bills and ten dollar bills in his pocket. the value of these bills is $100.00. how many f

ive dollar bills are in Lawrence's
pocket?
Mathematics
2 answers:
denis23 [38]3 years ago
5 0
10 $5 bills
5 $10 bills
erma4kov [3.2K]3 years ago
5 0

Answer:

There are 10 five dollars bills in Lawrence's  pocket.

Step-by-step explanation:

Given,

Total bills = 15,

Let x be the number of five dollar bills,

⇒ The number of ten dollars bill = 15 - x

Thus, the total amount = 5x + 10(15-x)

According to the question,

5x + 10(15-x) = 100

⇒ 5x + 150 - 10x = 100

⇒ - 5x + 150 = 100

⇒ -5x = 100 - 150

⇒ -5x = -50

⇒ x = 10

Hence, there are 10 five dollars bills in Lawrence's  pocket.

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16% of 150 is what number plz show work
bazaltina [42]
To get the solution, we are looking for, we need to point out what we know. 

<span>1. We assume, that the number 150 is 100% - because it's the output value of the task. </span>
<span>2. We assume, that x is the value we are looking for. </span>
<span>3. If 150 is 100%, so we can write it down as 150=100%. </span>
<span>4. We know, that x is 16% of the output value, so we can write it down as x=16%. </span>
5. Now we have two simple equations:
1) 150=100%
2) x=16%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
150/x=100%/16%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 16% of 150

150/x=100/16
<span>(150/x)*x=(100/16)*x       - </span>we multiply both sides of the equation by x
<span>150=6.25*x       - </span>we divide both sides of the equation by (6.25) to get x
<span>150/6.25=x </span>
<span>24=x </span>
x=24

<span>now we have: </span>
<span>16% of 150=24</span>
4 0
3 years ago
Read 2 more answers
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

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3 years ago
Use the x-intercept method to find all real solutions of the equation. -9x^3-72x^2+36=3x^3+x^2-3x+8
larisa86 [58]

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-72x^2 + 36 = 3x^3 + 9x^3 + x^2 - 3x + 8       Add 72x^2 to both sides

36 = 12x^3 +   73x^2 - 3x + 8                           Subtract 36 from both sides.

0 = 12x^3 + 73x^2 - 3x - 28      

It does factor, but it is not very nice.

(x + 6.06)(x - 6.09)(x + 0.632)

If there is any kind of error please report it in a note below.

6 0
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if rounded to the nearest dime what is the greatest amount of money that rounds to $105.40? what is the least amount of money th
Kamila [148]
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4 0
3 years ago
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galben [10]

Answer:

1st - c

2nd - d

Step-by-step explanation:

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