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Lera25 [3.4K]
2 years ago
6

The sum of four consecutive integers is 190 what is the third integer

Mathematics
1 answer:
patriot [66]2 years ago
6 0
X+x+1+x+2+x+3=190
4x+6=190
4x=184
x=46

the numbers are 46,47,48,49 so your answer is
48
You might be interested in
Francine Flicka found a bargain when she bought a lawn mower. She paid the store's clerk three $20 bills and received him $7.45
adoni [48]

Answer:

  $48.97

Step-by-step explanation:

Let p represent the marked price of the mower. Then the price with tax is ...

  with tax = p + 7.3%×p = 1.073p

The amount Flicka paid is the amount tendered less the change she received, so is ...

  3×$20 - 7.45 = with tax = 1.073p

To find p, we can divide by its coefficient:

  (3 × $20 - 7.45)/1.073 = p ≈ $48.97

The selling price before taxes was $48.97.

_____

<em>Comment on the problem</em>

If you check the answer, you find the amount with tax is ...

  $48.97 × 1.073 ≈ $52.54481 ≈ $52.54

so the change Flicka would have received would have been $7.46, not $7.45. If the price were $48.98, then the change would have been $7.44. There is no price at which the mower can be marked that will make the total with tax come to $52.55. There is no solution to this problem.

5 0
3 years ago
Read 2 more answers
pls help. Natalia paid $38.95 for three medium-sized pizzas and a salad. If Natalia paid $11.98 for the salad, how much did each
blagie [28]
38.95-11.98=26.97
26.97/3=$8.99
3 0
3 years ago
2. Which of the following ordered pairs (x,y) satisfies the system of equations below?
bixtya [17]

Answer:

b(2,-2)

Step-by-step explanation:

x= 2 and y=-2

let me know if you want further eplanation

3 0
2 years ago
AD←→ is tangent to circle M at point D. The measure of ∠DMQ is 58º.
Marat540 [252]

Answer:

32°

Step-by-step explanation:

Given:

∠DMQ = 58º

In this circle, the radius is DM. Since AD is tangent to the circle M, at point D, and the angle between a tangent and a radius is 90°

Therefore, ∠MDQ = 90°

The total angle in a triangle is 180°. Since we have the values of ∠MDQ and ∠DMQ, ∠DQM will be calculated as:

180 = ∠DMQ + ∠MDQ + ∠DQM

Solving for ∠DQM, we have:

∠DQM = 180 - ∠DMQ - ∠MDQ

∠DQM = 180 - 90 - 58

∠DQM = 32°

The measure of ∠DQM is 32°

6 0
2 years ago
An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0
3 years ago
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