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3 years ago

I need help on both of these questions 5/12 + 3/5 and 7/9 - 1/2 please

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

3 0

1) u would have to find the lcd of both. And that is 60 so it would be 25/60 +36/60 = 61/60 = 1 1/60

2) do the same thing as in number 1. Find the LCD it is 18. So 14/18 - 9/18 = 5/18

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For $7.52, you purchased 8 pens and highlighters from a local bookstore. Each highlighter cost $1.09 and each pen cost $.69. How

seraphim [82]

5 0

3 years ago

Use the data table provided to calculated the values requested below. Provide all answers to three decimal places.

babymother [125]

Answer:

1) 0.700

2) 0.730

3) 0.030

4) 0.959

Step-by-step explanation:

1) proportion of support for the ban with at least one child =

$\frac{no of support atleast 1 child}{Total no of atleast 1 child\\}$

= $\frac{1739}{2485}$

= 0.700

2) proportion of support for the ban with no child =

= $\frac{no of support with no child}{Total of no child}$

= $\frac{3089}{4231}$

= 0.730

3) Difference in proportion of supporters for the ban between those with atleast one child and those with no child

= 0.700 - 0.730

= -0.03

4) Relative risk = $\frac{proportion with atleast on child}{proportion with no child}$

= $\frac{0.700}{0.730}$ = 0.959

5 0

3 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

3 years ago

Can someone please help me

IrinaK [193]

For the mean, add together all of the numbers together (let's call the total x) and count how many numbers there are. (let's call this number y). divide x by y, and that is your mean. for the median, put all the numbers in order, and the median is the number in the middle. if there are two numbers in the middle, add them together and divide them by 2.

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3 years ago

Which equation is equivalent to ax+b=c

alexgriva [62]

Answer: b = c-ax

hope this helps:)

7 0

3 years ago

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