Write a decimal equation that has a product of 3.15

The sum of 5 consecutive odd integers is 2,555. What is the sum of the first, third, and last integer?

Lilit [14]

Answer:

The sum of the first, third, and last integer is 1533.

Step-by-step explanation:

Step 1: You need to create a equation first:

n + n + 2 + n + 4 + n + 6 + n + 8 = 2555

How did I make the equation?

If n is odd number, n + 2 must be the consecutive odd number

5 , 5 + 2 , 5 + 4 , 5 + 6 , 5 + 8

5, 7, 9, 11, 13 ALL CONSECUTIVE ODD NUMBERS

Step 2: Simplifying the equation

n + n + 2 + n + 4 + n + 6 + n + 8 = 2555

5n + 20 = 2555

Step 3: Solving the equation, with explanation

5n + 20 = 2555 Subtract 20 on both sides

5n = 2535 Isolate the variable by dividing 5 on both sides

n = 507

Step 4: Plug in the value

n + n + 2 + n + 4 + n + 6 + n + 8

507, 507 + 2 , 507 + 4 , 507 + 6 , 507 + 8

507, 509, 511, 513, 515

Step 5: Solve the sum of the first, third, and last integer

507 + 511 + 515 = 1533

Step 6: THE ANSWER

1533

<h3>Hope this helps!!! </h3><h3>Please mark this as brainliest!!! </h3><h3>Thank You!!! </h3><h3>:)</h3>

8 0

3 years ago

NAME THE LARGEST ANGLE NAME THE SMALLEST ANGLE

Natasha2012 [34]

Answer:

largest: cu smallest: rc

5 0

2 years ago

Read 2 more answers

In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2

Ne4ueva [31]

Answer:

The probability that on any given day the water supply is inadequate $0.1991$

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

$P(X>9)= 1-P(X\leq 9)\\$

Expanding the function and putting the given values, we get -

$[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991$

The probability that on any given day the water supply is inadequate $0.1991$

5 0

3 years ago

Find the critical value tc for the confidence level cequals 0.95 and sample size nequals 19.

Alex_Xolod [135]

The critical values are fixed values which can be obtained from the standard t crit tables. We can find the critical value tc when the degrees of freedom and confidence level are given. The degrees of freedom is simply:

Degrees of Freedom = Number of samples – 1

Degrees of Freedom = 19 – 1

Degrees of Freedom = 18

From the t critical tables, the tc corresponding to Degrees of Freedom equal to 18 and Confidence level equal to 95% is as follows:

tc = 2.101

Therefore this means that any t value beyond 2.101, we already reject the null hypothesis.

8 0

3 years ago

A company sells widgets. The amount of profit, y, made by the company, is related to the selling price of each widget, x, by the

Alex17521 [72]

The equation is that of a parabola, meaning that the maximum (or sometimes minimum) is 1/2 the sum of the solutions found using the quadratic equation.

y = 0 for the 2 solutions to {-b +/- sqrt[b2 - 4ac]} / 2a, where a = -3, b = 152 and c = -1150 from the equation for y. Plug a, b and c into the quadratic formula to get x = 9.2571 and x = 41.4095 as the 2 solutions.

9.2571 + 41.4095 = 50.6666... and the midpoint is 25.3333...

When x = 25.3333..., y = 775.33333.....

To the nearest cent, x=25.33 results in a profit of 775.33.

When x=25.25, Y = 775.31

When x=25.4, Y = 775.32

These are both less than 775.33 when x = 25.33.

3 0

3 years ago