there are 14,240 books in the library. they are arranged on shelves that hold 8 books each. How many shelves are i the library

A bulletin board can be covered completely by 30 square pieces of paper without any gaps or overlaps if each piece of paper has

34kurt

Answer:

The area of the bulletin board is 30 square foot.

Step-by-step explanation:

A bulletin board can be covered completely by 30 square pieces of paper without any gaps or overlaps.

If each piece of paper has side length of 1 foot so the area of each paper is 1 ft x 1 ft = 1 square foot.

There are 30 pieces of paper needed to completely cover the bulletin board, so 30 pcs of paper with an area of 1 square foot is equal to 30 square feet. The area of the bulletin board is 30 square foot.

4 0

2 years ago

What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (–5, 1)? Check all that ap

Bond [772]

So the right options are:

$y=-\frac{2}{5}x-1$

$2x+5y = -5$

$y-1=-\frac{2}{5}(x+5)$

Further explanation:

Given equation of line is:

2x+5y=10

We have to convert it into point-slope form

$2x+5y=10\\5y=-2x+10\\Dividing\ both\ sides\ by\ 5\\y=-\frac{2}{5}x+\frac{10}{5}\\y=-\frac{2}{5}x+2$

The co-efficient of x is the slope of the line

So,

$m= -\frac{2}{5}$

As the required line is parallel to given line, it will also have same slope.

Let m1 be the slope of required line

Then the line will be:

$y=m_1x+b$

Putting the value of slope

$y=\frac{2}{5}x+b$

Putting (-5,1) in the equation to find the value of b

$1=-\frac{2}{5}(-5)+b\\1=2+b\\b=1-2\\b=-1$

Putting the values of slope and b in equation

$y=-\frac{2}{5}x-1$

Multiplying the whole equation by 5 will give us:

5y = -2x-5

2x+5y = -5

Another form of equation of line is Point-slope form

$y-y_1=m(x-x_1)$

Putting the values of slope and point in the equation, we get

$y-1=-\frac{2}{5}(x+5)$

So the right options are:

$y=-\frac{2}{5}x-1$

$2x+5y = -5$

$y-1=-\frac{2}{5}(x+5)$

Keywords: Point-Slope form, Parallel lines

Learn more about point slope form at:

#LearnwithBrainly

4 0

2 years ago

Read 2 more answers

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

A retail store offers all television sets at 80%, or , of their original price if the customer buys a store membership for $100.

Natasha_Volkova [10]

Let X = be the amount the amount that new customer will have to pay X = 0.8P + 100 Given that the television is $1500, the discounted price is $1200. But the customer will have to pay the membership of $100. Therefore the total amount is $1300.

8 0

2 years ago

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The hypotenuse of right triangle ABC, line segment AC, measures 13 cm. The length of line segment BC is 5 cm.

slavikrds [6]

Cos ( ∠C ) = 5/13 = 0.38462
m ∠C = cos^(-1) 0.38642 = 67.4°
m ∠A = 90° - 67.4° = 22.6°
m ∠C - m ∠A = 67.4° - 22.6° = 44.8°
Answer:
B ) 44.8°

7 0

2 years ago