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max2010maxim [7]

2 years ago

6

Amy loaned \$5$5dollar sign, 5 to each of 202020 people. She recorded each loan by writing -\$5−$5minus, dollar sign, 5 in her b

ank book. What is the total change in the amount of money Amy had before the loans?

1 answer:

polet [3.4K]2 years ago

7 0

Answer:

-$100

Step-by-step explanation:

If Amy subtracted $5 from her bank balance 20 times, it is ...

$5 × 20 = $100

<em>less</em> than it was.

The amount of change totaled -$100.

You might be interested in

Which shows how to solve the equation Three-fourths x = negative 6 for x in one step?

My name is Ann [436]

option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).

<u>Step-by-step explanation:</u>

We have , The given expression as Three-fourths x = negative 6 , which can be written as $\frac{3}{4} (x) = -6$ . Now in order to solve this equation in one step , we must notice that coefficient of x must be 1 but it's $\frac{3}{4}$ , Let's make coefficient of x as 1 by multiplying both side of equations by 4/3:

$\frac{3}{4} (x) = -6$

⇒ $\frac{3}{4} (x) = -6$

⇒ $\frac{3}{4}(\frac{4}{3} ) (x) = -6(\frac{4}{3} )$

⇒ $x = -6(\frac{4}{3} )$

⇒ $x = -2(4)$

⇒ $x = -8$

Therefore, x= -8 & correct option to solve the equation Three-fourths x = negative 6 for x in one step is <u>option A i.e. Three-fourths (four-thirds) x = negative 6 (four-thirds).</u>

7 0

2 years ago

Read 2 more answers

Tyler had $65.40 in his checking account. Then he wrote checks on the account for $20.38, $11.48, and $19.50. What is the balanc

andrew-mc [135]

money going out (20.38+11.48+19.50) =51.36

balance - out = new balance

65.40-51.36 =

14.04

new balance = 14.04

4 0

2 years ago

Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t

musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago

What is the lateral area of a right rectangular prism that has rectangular bases measuring 3" by 5" and a height of 6"?

Mademuasel [1]

Answer:

The answer is the option A

$96\ in^{2}$

Step-by-step explanation:

we know that

The lateral area of a right rectangular prism is equal to

$LA=P*h$

where

P is the perimeter of the base

h is the height of the prism

Find the perimeter

$P=2*(3+5)=16\ in$

$h=6\ in$

substitute in the formula

$LA=16*6=96\ in^{2}$

6 0

2 years ago

Read 2 more answers