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4 years ago

What is 1+1=? Plz help

Mathematics

2 answers:

DanielleElmas [232]4 years ago

3 0

Answer: 2

Step-by-step explanation:

SVEN [57.7K]4 years ago

3 0

Answer:

Step-by-step explanation:

2=1+1

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What is the avarege of -2.5,5.2,1.7 and -0.8

aleksandr82 [10.1K]

Average:

4.4 / 3 = 1.4666666666667

Count:

Sum:

4.4

Average:

4.4 / 3 = 1.4666666666667

Count:

Sum:

4.4

5 0

3 years ago

1.75 + .75m = 4.75 what’s the answer in solution for this problem

ELEN [110]

Answer:

m=4

Step-by-step explanation:

We are to find the value of m in the question

1.75+0.75m=4.75

Let's start by substrating 1.75 from both sides

0.75m=3

We will have to make m the subject of formula by dividing both sides by 0.75

m=4

Therefore the final answer for m is 4

5 0

3 years ago

Read 2 more answers

Help please 10 points

katovenus [111]

Answer:

The answer would be D - (11, -1)

Step-by-step explanation:

You can this by adding 6 to the x-coordinate since it is shifting right and then subtracting 4 units from the y-coordinate.

6 0

3 years ago

Read 2 more answers

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

6 0

3 years ago

Damage cost (x) to a car in a crash is modeled by c(x2 – 60x + 800) , 0 < x < 20 f (x) = 0, otherwise where c is a constan

mixas84 [53]

Answer:

0.2

Step-by-step explanation:

solution is down below

7 0

4 years ago