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2 years ago

L= a+(n-1)d SOLVE FOR N

Mathematics

1 answer:

Deffense [45]2 years ago

8 0

Answer:

$n = \frac{l-a}{d}+1$

Step-by-step explanation:

$l= a+(n-1)d\\$

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

Divide both sides of the equation by d

$\frac{l-a}{d} = \frac{(n-1)}{d}\\ \\\frac{l-a}{d} = n-1\\\\\frac{l-a}{d}+1 = n$

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Georgia [21]

Answer:

Solution-

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Step-by-step explanation:

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2 years ago

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Consider the graph of Bx + Cy = A where A, B, and C are all positive constants. Find the coordinates of the x-intercept. Use upp

timofeeve [1]

That is where the line crosses the x axis or where y=0
set y=0
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2 years ago

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One morning, eight buses arrive at a bus stop.

Semenov [28]

Answer:

0 and 5 minutes

Step-by-step explanation:

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

We have given 8 numbers and we have to find the other two numbers a minute late for each bus to arrive at the bus stop.

Also, we have given the median and mode of all 10 buses i.e. 3.5 and 0 respectively.

Since the mode is 0, so the frequency of 0 must be greater than 2 times and less than or equal to 4 times.

The frequency of 0 is 4 is not possible as then we didn't get a median 3.5 then.

So the frequency of 0 is 3.

Hence one of number is 0 in two unknown number of a minute late for each bus arrive at the bus stop.

Now, we have the number of minutes late for each bus after arranging them in ascending order is: 0 0 0 2 2 6 8 8 9

For getting a median 3.5 we must have a number between 2 and 6. Let it be x then:

$\frac{2+x}{2} = 3.5$

⇒x = 5

Hence the number of minutes late for each bus is: 0 0 0 2 2 5 6 8 8 9

Thus, 0 and 5 are minutes late for the two-afternoon buses.

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2 years ago

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Five light bulbs that are either on or off are lined up in a row. A certain code uses different combinations of lights turned on

Paraphin [41]

Answer:

Step-by-step explanation:

The number of different ways that 2 lights can be chosen from 5 is ₅C₂. Use a calculator or find the value manually:

₅C₂ = 5! / (2! (5−2)!)

₅C₂ = 5! / (2! 3!)

₅C₂ = 5×4×3×2×1 / (2×1 × 3×2×1)

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2 years ago

The range of all real numbers x such that 2x − 5 < 7 OR 4x + 10 > 6 is___

scoray [572]

First, simplify the two inequalities.

2x - 5 < 7
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4x + 10 > 6
4x > -4 <-- Subtract both sides by 10
x > -1 <-- Divide both sides 4

Now we have the compound inequality x < 6 OR x > -1

To fulfill the first inequality, x has to be less 6. For the second inequality, x has to bigger than -1.

So, the answer is B. (-infinity, infinity).

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2 years ago

L= a+(n-1)d SOLVE FOR N ​

L= a+(n-1)d SOLVE FOR N