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mezya [45]
3 years ago
6

Is 3 greater than -2

Mathematics
2 answers:
dimaraw [331]3 years ago
3 0
Yes, three is greater than -2
astraxan [27]3 years ago
3 0
Yes three is greater than -2. If you compare these two numbers on a number line than three will always be greater. Positive numbers are always greater than negative numbers
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Negative. (Zero) positive

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Joel is a service assistant at a supermarket. He is 19 years old.<br><br>What is Joel’s weekly pay?
Talja [164]

Answer:

$452.72

Step-by-step explanation:

Joel is 19 and according to the chart he makes 80% of what an adult over 21 makes

$565.90 * 80% = $452.72

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Which statements prove that a quadrilateral is a parallelogram?
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The answers are A, B and D.
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Solve 2x - 1 &lt; 7 and 5x + 3 &lt; 3. {x | x &lt; 0} {x | x &lt; 4} {x | 0 &lt; x &lt; 4}
nataly862011 [7]
2x -1 <7 ; then, 2x < 8; then x < 4;
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A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She esti
Alona [7]

Answer:

(a) <em>X</em> = number of dealerships the customer needs to call before she finds a used red Miata car.

(b) <em>X</em> = 1, 2, 3, 4,...

(c) The distribution of <em>X</em> is, X\sim Geometric\ (p=0.28).

(d) The expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e) The probability that she must call at most 4 dealerships is 0.7313.

(f) The probability that she must call 3 or 4 dealerships is 0.2497.

Step-by-step explanation:

(a)

The random variable <em>X</em> can be defined as the number of dealerships the customer needs to call before she finds a used red Miata car.

(b)

The values that the random variable <em>X</em> can assume are:

<em>X</em> = 1, 2, 3, 4,...

(c)

The random variable <em>X</em> is defined as the number of trials before the first success. Each trial is independent of the others. And each trial has a similar probability of success.

This implies that the random variable <em>X</em> follows a Geometric distribution.

The probability of success, i.e. the probability that any independent dealership will have the car is <em>p</em> = 0.28.

Thus, the distribution of <em>X</em> is, X\sim Geometric\ (p=0.28).

(d)

The expected value of a Geometric distribution is:

E(X)=\frac{1}{p}

Compute the expected number of dealerships would we expect her to call until she finds one that has the car as follows:

E(X)=\frac{1}{p}

         =\frac{1}{0.28}\\

         =3.57

Thus, the expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e)

Compute the probability that she must call at most 4 dealerships as follows:

P (X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

              =\sum\limits^{4}_{x=1}{(1-0.28)^{x-1}\times 0.28}\\=0.28+0.2016+0.1452+0.1045\\=0.7313

Thus, the probability that she must call at most 4 dealerships is 0.7313.

(f)

Compute the probability that she must call 3 or 4 dealerships as follows:

P (X = 3 or X = 4) = P (X = 3) + P (X = 4)

                            =[(1-0.28)^{2-1}\times 0.28]+[(1-0.28)^{4-1}\times 0.28]\\=0.14515+0.10451\\=0.24966\\\approx0.2497

Thus, the probability that she must call 3 or 4 dealerships is 0.2497.

6 0
3 years ago
I need help, Gonna give brainliest to right answer
Brums [2.3K]

Answer:

s = 7t

Step-by-step explanation:

8 0
2 years ago
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