There were 400 students in attendance. 51% are boys. How many are boys?

Use PEMDAS to simplify the expression.<br> 7 715-221<br> 2<br> + 10 - (5 x 23)

Bogdan [553]

Answer:

5398

Step-by-step explanation:

7715-2212+10-(5x23)

7715-2212+10-115

5503+10-115

5513-115

5398

3 0

3 years ago

Explain how you can use base 10 blocks to find 2.16÷3

Vikentia [17]

You can use 10 blocks by having 1 ten and 6 ones and then divide it into groups of 3 when you do that you are kind of subtracting when you take away 3 away each time. After that you multiply it, so you think of whatever times 3 is 16 which is 15 r1. Then your answer should be 30.

8 0

3 years ago

PLEASE HELP ME ITS DUE IN 1 HOUR

Vadim26 [7]

Answer: Answer:Median: 10.5Lower quartile: 4Upper quartile: 16Explanation:Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 12 (even)Median = Mean of (6)th and (7)th observations = 10 + 11 divided by 2 = 21/2 = 10.5Therefore the median is 10.5.After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively.They are:Lower quartile: 4Upper quartile: 16Question 2.the lengths of 10 pencils, in centimeters:18, 15, 4, 9, 14, 17, 16, 6, 8, 10Type below:_________________Answer:Median: 12Lower quartile: 4Upper quartile: 18Explanation:Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 10 (even)Median = Mean of (5)th and (6)th observations = 10 + 14 divided by 2 = 24/2 = 12Therefore the median is 12.After writing in ascending or descending order the first and last terms justify the lower and upper limits respectively.They are:Lower quartile: 4Upper quartile: 18Question 3.Make a box plot to display the data set in Exercise 2.Type below:_________________Answer:The above box plot represents the lower and upper quartiles, the median.Explanation:Box plot is drawn using the number line and the rectangle which is drawn above it.The ends of the rectangles say about the lower and upper limits and the middle line indicates the median.Question 4.The numbers of students on several teams are 9, 4, 5, 10, 11, 9, 8, and 6. Make a box plot for the data.Type below:_________________Answer: Explanation:Box plot is drawn using the number line and the rectangle which is drawn above it.The ends of the rectangles say about the lower and upper limits and the middle line indicates the median.Therefore the lower and upper quartiles are 4 and 11 respectively.Median:First, write the observations in ascending order or descending order.The formula to calculate the median is mean of (n/2) and (n/2+1)th observations, for even number of observationsn = 8 (even)Median = Mean of (4)th and (5)th observations = 8 + 9 divided by 2 = 17/2 = 8.5Therefore the median is 8.5.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h$

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

What is the completely factored form of this polynomial?<br> 18x^3– 120x^2-42x

Vlada [557]

Answer:

6×(3×+1)(×-7)

Step-by-step explanation:

18×^3-120×^2-42×

=6×(3×+1)(×-7)

4 0

3 years ago

Read 2 more answers