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maks197457 [2]
3 years ago
12

on a random Monday in the spring, 95% of students are present at school. if there are 589 students present, then how many studen

ts are enrolled?​
Mathematics
1 answer:
Rus_ich [418]3 years ago
6 0

Answer:

620

Step-by-step explanation:

95% × 620 =

(95 ÷ 100) × 620 =

(95 × 620) ÷ 100 =

58,900 ÷ 100 =

589;

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Allen rents a truck for 45 dollars plus .36 cents for every mile that he drives the truck. Write an equation that models this li
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Answer:

This is a linear equation of the form y = mx + b, where m is the rate of change and b is the y-intercept, also known as the value of y when x = 0.  Our independent variable is the number of miles we travel, and we will call that x.  If we pay .20 per mile, and miles is "x", we represent that as .20x.  The 70 is the amount we will pay per day even if we drive 0 miles.  Therefore, the equation is:

y = .20x + 70.

I cannot graph this for you here.

We can choose a number of miles, say 100, and solve for our cost:

y = .20(100) + 70

y = 20 + 70

y = 90

We can expect to pay $90 if we drive 100 miles in a day.

Read more on Brainly.com - brainly.com/question/13029200#readmore

Step-by-step explanation:

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Hope this helps! :)

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Step-by-step explanation:

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3 years ago
Find the slope of the line whose equation is 0.05x - 0.03y = 9.
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Using the slope-intercept form (y=mx+b) the slope is:
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5 0
3 years ago
Read 2 more answers
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

8 0
2 years ago
Read 2 more answers
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