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Alinara [238K]
3 years ago
5

Taylor surveys students in one grade level who own at least one pet. She finds that 50% of the students surveyed own 2 pets, 3 s

tudents own 3 pets each, and 2 students own 4 pets each. Eight of the students in the grade own 1 pet. Considering the number of pets as the random variable, X, which of the following is the probability distribution, PX(x)? A probability distribution is shown. The probability of 1 is 0.3; 2 is 0.5; 3 is 0.11; 4 is 0.08. A probability distribution is shown. The probability of 1 is 0.28; 2 is 0.5; 3 is 0.1; 4 is 0.13. A probability distribution is shown. The probability of 1 is 0.61; 2 is 0.5; 3 is 0.22; 4 is 0.15. A probability distribution is shown. The probability of 1 is 0.53; 2 is 0.5; 3 is 0.2; 4 is 0.27.
Mathematics
2 answers:
Shalnov [3]3 years ago
8 0

First, we need to work out the total number of students who were being surveyed.

We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population

That means total number of students being surveyed is 13+13=26 students

Then we work out the probability

P(One pet) = 8/26 = 4/13

P(Two pets) = 1/2

P(Three pets) = 3/26

P( Four pets) = 2/26 = 1/13

The probability distribution is shown in the table below. Let  be the number of pets and  is the probability of owning the number of pets

Rasek [7]3 years ago
7 0

Answer:

Step-by-step explanation:

We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population

That means total number of students being surveyed is 13+13=26 students

Then we work out the probability

P(One pet) = 8/26 = 4/13

P(Two pets) = 1/2

P(Three pets) = 3/26

P( Four pets) = 2/26 = 1/13

The probability distribution is shown in the table below. Let  be the number of pets and  is the probability of owning the number of pets

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In a genetics experiment on peas, one sample of offspring contained green peas and yellow peas. Based on those results, estimate
dem82 [27]

Complete Question

In a genetic experiment on peas, one sample of offspring contained 436 green peas and 171 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the value of 3/4 that was expected? The probability of getting a green pea is approximately: Is the probability reasonably close to 3/4?

Answer:

The  probability is  P(g) =0.72

Yes the result is reasonably close

Step-by-step explanation:

From the question we are told that

   The  number of  of  green peas is  g =  436

     The number of yellow peas is  y  = 171

   The sample size is  n  =  171 +  436 =  607

The probability of getting an offspring pea that is green is mathematically represented as

       P(g) = \frac{g}{n}

        P(g) = \frac{436}{607}

        P(g) =0.72

Comparing  P(g) =0.72  to   \frac{3}{4}  =  0.75 we see that the result is reasonably close

7 0
3 years ago
6. Find the function with x-intercepts (-3,0) and (5,0), which also goes through the point (1,8)
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Answer:

f(x) = (-1/2)(x^2 + 8x - 15)

Step-by-step explanation:

This function has two roots:  -3 and 5.  Most likely it is a quadratic (all of which have two roots).

Then f(x) = a(x + 3)(x - 5)

The graph goes through (1. 8):  Therefore, y = 8 when x = 1:

f(1) = a(1 + 3)(1 - 5) = 8, or

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            -16a = 8, which leads to a = -1/2.

Thus the quadratic in question is f(x) = (-1/2)(x + 3)(x - 5), or

                                                        f(x) = (-1/2)(x^2 + 8x - 15)

4 0
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Ivenika [448]

Answer:

Y=-2x

When x= -6 then y =12

Step-by-step explanation:

If y and x in direct variation then y=c*x for some constant c.

We are told

-4=c2

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So y= -2x

When x= -6 this becomes

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8 0
3 years ago
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